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SCHOOL OF EDUCATION
LIBRARY
WHITE'S SERIES OF MATHEMATICS
A
SCHOOL ALGEBRA
DESIGNED FOR tJSE IN
siair SCHOOLS and academies
BY
EMERSON E. J^HITE, A.M., LL.D.
AUTHOH OF "SEBIES OF MATHEMATICS," '* ELEMENTS OF PeDAGOGT,**
" School Management," xto.
>;vi»
NEW YORK . : • CINCINNATI . : • CHICAGO
AMERICAN BOOK COMPANY
WHITE'S MATHEMATICS
Oral Lessons in Number
(for teachers)
First Book of Arithmetic
New Complete Arithmetic
School Algebra
Elements of Geometry
WHITE'S PEDAGOGICS
Elements of Pedagogy
School Management
WHITE'S SCHOOL RECORDS
New School Register
Monthly School Record
Teachers' Class Record
COPTBIOHT, 1896, BT AXXKIOAK BoOK GOMFAlfT.
I P7
PREFACE.
•c*
This work is designed for use in high schools and academies, and it
covers sufficient ground to meet fully the entrance requirements of the
best colleges and universities. It is specially adapted to the first two
years of the usual highnschool course.
It has been the author's aim to prepare a school algebra that is
pedagogically sound, as well as mathematically accurate, and thoroughly
adequate for its place and purpose. He has kept in mind the fact
that the great majority of the pupils who begin the study of algebra
are between thirteen and fifteen years of age, and hence are too young
and immature to master successfully a textbook designed for older and
more advanced students. These young pupils have as a class a fair
knowledge of the analytic and inductive processes of arithmetic; and
true pedagogical principles require that this prior training be made as
helpful as possible in their introduction to algebra.
The manner in which these ends have been attained is partially
indicated by the following statements : —
1. The early introduction and practical use of the equation. The equa
tion, properly called *Uhe instrument of algebra," is used from the
beginning in the solution of simple problems, thus awakening at the
first the pupil's interest in the new study. The pupil has been using
the informal equation throughout his entire course in arithmetic, and so
finds no difficulty in using the equation in solving problems in advance
of its formal treatment. The equation is also freely used in proofs and
explanations, and in establishing data for inductions and other generali
zations. This use of the equation avoids those long and abstruse verbal
statements which so baffle the young learner.
2. The use of arithmetical approaches to algebraic processes and princi
ples, especially in the first half of the work. Since arithmetic deals with
particular numbers, and algebra with both particular and general num
bers, the processes and principles of arithmetic afford a natural and easy
approach to those of algebra. This conforms to two fundamental princi
ples of procedure in elementary teaching ; to wit, ^^from the particular
to the general,"*^ and ^^from the known to the related unknown.''^ These
pedagogical principles apply to the teaching of the elements of all science,
and they are specially helpful in the pupil's introduction to algebra.
3. The intelligent use of the inductive method. The introduction of the
inductive method is a characteristic feature of modern arithmetic, and its
increasing use in algebra establishes a still closer relation between the two
studies. In the present work, the inductive method is used, when practica
3
4 PREFACE.
ble, and algebraic principles and laws are thus easily and clearly reached.
When formal rules are given, they are placed ajfter the exercises and
problems, as in the author*s ** Series of Arithmetics,** — an arrangement
specially commended by the "Committee of Ten." The method of
deductive proof is introduced gradually as the pupil advances towards
the closing chapters, thus making him somewhat familiar with the method
of demonstration which he will subsequently use in geometry.
4. The immediate application of facta and principles in simple exercises
for practice; this being a marked feature of the chapter on algebraic
notation, often so difficult for beginners. Practice exercises not only
help fix the facts and principles taught in the memory, but they greatly
increase the clearness with which they are comprehended by the young
learner. A principle that may not be fully understood from its state
ment is often clearly grasped in its application.
Attention is specially called to the character of the exercises and prob
lems. The essential result to be attained by the pupil in the study of the
elements of algebra is facility and accuracy in algebraic processes.
This is necessary to all satisfactory future progress, and, to this end, there
must be abundant and appropriate practice. Great pains have been taken
to make the exercises and problems in this work adequate in number^
variety^ and grading. There has been a careful exclusion of problems
believed to be too difficult. Such problems not only discourage the
learner, but call for an unprofitable use of time and energy.
Progressive teachers will be pleased to find a number of subjects and
processes not given in the school algebras in general use. There will
also be found throughout the work new and elegant solutions, and
other new features of special interest and value. It is believed that
few textbooks have been prepared with greater care, or v^ith more
earnest effort to ascertain and meet the needs of the schools. The
result is a progressive modern algebra.
The author gratefully acknowledges his indebtedness to Professor
John Macnie, of the University of North Dakota, for many exercises
and problems, and for other contributions of subjectmatter ; to Professor
M. C. Stevens, in charge of the department of mathematics in Purdue
University, Indiana, for a critical reading of the manuscript, and for
many helpful suggestions, including new solutions and proofs, and other
material ; to Professor E. A, Lyman, of the University of Michigan, for
the critical reading of the manuscript, valuable suggestions, and other
help ; and also to several experienced teachers of algebra in high schools,
who have rendered important assistance.
Columbus, O.
CONTENTS.
aiAPTSIt
PAOB
L
Introduction • • • 7
Algebraic Equations . • • • <
. 7
n.
Algebraic Notation . • • • <
. 16
Positive and Negative Numbers . •
. 30
Tiaws ,
. 33
Equations and Problems . . « «
. 39
m.
Addition and Subtraction . . • «
43
IV.
Multiplication and Division . • • «
Fractional Coefficients
54
. 67
Detached Coefficients .
> . • • «
. 68
Synthetic Division
> • • • 4
71
V.
Simple Equations .
• . • «
74
VL
Formulas ....
> • . . «
^
85
Special Fonns in Multiplication . . .
85
Division by Binomial Factors
. 90
vu.
Factoring
Special Methods of Factoring
General Method of Factoring Trinomials
Factoring by Synthetic Division .
95
95
108
113
vm.
Common Factors and Multiples .
119
IX.
Fractions ,
Reduction of Fractions
Addition and Subtraction of Fractions .
Multiplication and Division of Fractions
130
132
139
143
X.
Simple Equations containing Fractions
162
XI.
Simultaneous Equations ....
> a
. 169
Simple Equations with Two Unknown Numbers .
. 169
Equations with Three or More Unknown Number
3
179
XM.
Involution and Evolution
. 186
Powers .
. 186
Roots . . . . r . . .
. 193
XIII.
Radicals
Reduction of Radicals .....
Addition and Subtraction of Radicals .  .
Multiplication of Radicals
. 209
. 210
. 215
. 216
Division of Radicals
. 219
6 CONTENTS.
CHAPTSB PAOS
Xin. Radicals (continued).
Involution and Evolution of Radicals • • • . 222
Equations involving Radicals 226
Imaginary Numbers 228
XIV. Fractional and Negative Exponents .... 234
XV. Quadratic Equations 239
Incomplete Quadratics 240
Complete Quadratics 243
Literal Quadratics 253
Equations Quadratic in Form 257
XVL Simultaneous Quadratic Equations .... 265
General Methods of Solution 265
Special Methods 269
XVII. Inequalities •. . . 273
XVIII. Ratio, Proportion, Variation 277
Ratio 277
Proportion 281
Variation 287
XIX. Progressions 293
Arithmetical Progression 293
Geometrical Progression 300
Harmonic Progression 307
XX. Logarithms 309
Principles 311
Table with Tabular Differences 319
Applications to Numerical Processes .... 322
XXI. Undetermined Coefficients and Applications . . 329
Resolution of Fractions 330
Expansion of Fractions into Series .... 332
Binomial Formula 334
XXIL Determinants 340
Determinants of the Third Order 343
Determinants of Any Order 347
Properties 350
XXIII. Curve Tracing 355
Geometrical Representation of Equations . . . 357
Geometrical Representation of the Roots of an Equation, 362
XXIV. Permutations and Combinations 364
Permutations 364
Combinations 367
Appendix 369
Answers . . . . 375
ALGEBRA.
CHAPTER L
INTRODUCTION.
ALGEBRAIC EQUATIONS.
1. If a denotes a certain number, 3 a will denote 3 times
the number ; 4 a, 4 times the number ; and so on.
1. If a denotes a number, what will denote 5 times the
number ? Seven times the number ?
2. If n denotes a number, what will 5 n denote ? 8 n ? 12 n ?
3. If a? denotes the number of feet in a rod, what will de
note the number of feet in 3 rods ? In 6 rods ? In 12 rods ?
4. If a; denotes the number of bushels of apples in a barrel,
what will denote the number of bushels in 5 barrels? In
15 barrels ? In 20 barrels ?
6. If a; denotes a man's age, what will denote 3 times his
age ? 8 times his age ? f of his age ?
2. In this introductory chapter, the signs +, — , x, 5, and
= have the same meaning and use as in arithmetic.
The expression 7 + 5 denotes that 5 is to be added to 7, and
7 — 5 denotes that 5 is to be subtracted from 7. In like manner,
a f 6 denotes that the number represented by h is to be added
to the number represented by a, and a — h denotes that the
7
g ALGEBRA. [§ 2.
number represented by 6 is to be subtracted from the number
represented by a.
6. The sum of 3 a and 2 a is expressed by 3 a + 2 a, which
is 5 a. In like manner express the sum of 5 a and 4 a. How
many times a in the sum ?
7. Express by the sign + the sum of 4® and 6x. How
many times x in the sum ?
8. Express the sum of 3 a;, 2 a;, and 5x, How many times
X in the sum ?
9. If a; denotes A's age, and Sx B's age, what will express
the sum of their ages ? How many times x in the sum ?
10. If x denotes the cost of a chair, 3'x the cost of a table,
and 5 x the cost of a lounge, what will express the cost of the
three articles ? How many times x in the cost ?
11. An orchard contains 5 rows of pear trees, 4 rows of
peach trees, and 7 rows of apple trees. If x denotes the num
ber of trees in each row, what will express the number of
trees in the orchard ? How many times x in the number of
trees ?
12. A number is divided into two parts such that the greater
part is 4 times the less. If x denotes the less part, what will
denote the greater ? What will express the sum of the two
parts ? How many times x in the number ?
13. The difference expressed by 5 a — 3 a is 2 a. What is
the difference expressed by 7 a — 4 a ? How many times a in
the difference ?
14. Whatisthedifferenceexpressedby 9a— 5a? 13a— 7a?
20a15a? 17a9a? 21a12a?
15. What is the difference expressed by7aj— 4a;? 8a;— 3a;?
12a;7a;? 15a;6a;? 21a;7a;?
16. If 5 a; denotes A's age, and 3 x denotes B's age, what will
express the difference of their ages ? How many times x years
19 A older than B ?
§ 8.] INTRODUCTION. 9
3. If 4aj = 20, 1 Xf or simply a?, is \ of 20, which is 5; and
if X equals 5, 3 a; is three times 5, which is 16. The number 6,
which X here denotes, is called the value of x,
17. If 3aj = 15, what is the value of a?? Of 5aj? Of 9aj?
18. If Sx\4:X = 35, what is the value of a?? Of 4a??
19. If 6 a? + 4 aj = 90, what is the value of a? ?
20. If 7a?H5a;f 3aj = 45, what is the value of aj?
21. If 12 a?  7 a? = 30, what is the value of a; ?
22. If 7 aj + 4 a? — 5 a; = 24, what is the value of a? ?
23. If 8 a? + 2 a; — 7 a? = 21, what is the value of »?
4. The expression 7 + 5 = 3x4 denotes that the sum of
7 and 5 is equal to the product of 3 and 4. In like manner,
a\ b = c X d denotes that the sum of the numbers represented
by a and b is equal to the product of the numbers represented
by c and d,
5. The equality of two numbers may be expressed by the
sign =, which is read "equals" or "is equal to." Thus,
2 a; + 3 aj = 25 is read " two x plus three x equals 25"
6. An expression denoting the equality of two numbers
is called an equation. Thus, 2aj+3a; = 25isan equation.
7. An equation in which all of the numbers are expressed
by figures is called an arithmetical equation. An equation in
which one or more of the numbers is expressed by letters is
called an algebraic equation. Thus, 8 x 5 — 16 = 6 x 4 is an
arithmetical equation ; and 2 a? + 5 = 15, a + 6 = 20, and
a — 6 = c, are algebraic equations.
8. The solution of problems by means of an algebraic
equation is called the algebraic method.
The first step in the solution of a problem by the algebraic
method is to state the conditions of the problem in the form
10 ALGEBRA. [§ 9.
of an equation^ and the second step is to find the value of the
unknown number.
The finding of the value of the unknown number in an
equation is called the solution of the equation.
9. The advantage of the algebraic method of solving prob
lems is best shown by its actual use in the solution of problems
which can also be readily solved by the methods of arithmetic.
Take, for example, this problem :
A's age is twice B's age, and the sum of their ages is 60
years. What is the age of each?
Arithmetical Solution.
By the conditions of the problem, A's age is twice B's age,
and B's age plus twice B's age, or 3 times B's age, is 60 years.
Hence B's age is one third of 60 years, which is 20 years;
and A's age is twice 20 years, which is 40 years.
Algebbaio Solution.
Let X denote B's age ; then 2 x will denote A's age, and, by
the conditions of the problem, we have
ajf 2a; = 60;
whence 3 a; = 60 ;
and X = 20, B's age ;
and 2 a? = 40, A's age.
Hence B's age is 20 years, and A's age is 40 years.
10. In the algebraic statement of a problem, each number is
considered as abstract. Thus, since 20 years = 20 x 1 year, x
in the above solution represents 20, the concrete unit (1 year)
being omitted. To express the number of years, the abstract
value of X, when found, is considered as multiplied by the
omitted concrete unit (1 year).
§ 10.] INTRODUCTION. 11
Problems.
1. A and B together have $45^ and A has twice as much
money as B. How much money has each ?
Abithmetical Solution.
Twice B's money = A's money ;
then twice B's money + B's money = 9 46.
Hence 3 times B^s money = $ 45 ;
whence B^s money =  of $ 45 = $ 15,
and A's money =2 x $ 15 = $ 30.
Algbbsaic Solution.
Let X = B's money ;
then 2 a; = A's money,
and a; + 2a; = 39; = B's and A's money.
Hence 3 a: = 45 ;
whence a; = J of 45 = 15,
and 2a; = 2x 15 = 30.
Hence A has $ 30, and B $ 15.
Solve the next eight problems first arithmetically and then
algebraically.
2. A father and his son earn together $ 56 a month, and the
father earns 3 times as much as the son. How much does
each earn ?
3. A man paid f 24 for a coat and vest, and the coat cost
5 times as much as the vest. What was the cost of each ?
4. The sum of two numbers is 42, and the greater number
is 5 times the less. Find the numbers.
6. Divide $36 into two parts such that the greater shall
be 3 times the less.
6. Cut a piece of tape 30 yards long into two pieces such
thatlihe longer piece shall contain 5 times as many yards as
the shorter.
12 ALGEBRA. [§ 10.
7. A and B tc^tlier own 150 sheep, and A owns twice as
many sheep as B. How many sheep does each own ?
8. The sum of two numbers is 90, and 4 times the less
number equals the greater. . What are the numbers ?
9. If a number be increased by twice itself, the result will
be 60. What is the number ?
10. The difference between two numbers is 24, and the
greater number is 4 times the less. Wliat are the numbers?
Let X = the less number ;
then 4 X = the greater number,
and 4x — X = 3x = their difference.
Hence 3x = 24;
whence x = 8, the less number,
and 4x = 32, the greater number.
11. A father's age is 3 times the age of his son, and the
difference of their ages is 30 years. What is the age of each ?
12. Divide a number into two parts such that the greater
part will be 4 times the less, and their difference 81. What is
each part ? What is the number ?
13. Three times the cost of a saddle was the cost of a
harness, and the harness cost $ 12 more than the saddle. What
was the cost of each ?
14. A school enrolls 180 pupils, and there are 20 more girls
than boys. How many pupils of each sex in the school ?
Let X — number of boys ;
then X + 20 = number of girls,
and 2 X + 20 = number of pupils.
Hence 2x + 20 = 180.
Subtracting 20, 2 x = 160 ;
whence x = 80, number of boys ;
X + 20 = 100, number of girls.
If a number  6 = 35, then the number =35 — 5, which is 30. In
like manner if x + 6 = 35, then x = 35 — 5 = 30 ; and if x — 5 = 25, then
X = 25 + 5 = 30. It is thus seen that a number may be added to or sub
tracted from both members of an equation without affecting their equality.
§ 10.] INTRODUCTION. 18
16. A pole 120 feet long fell and broke into two pieces, one
piece being 30 feet longer than the other. What was the
length of each piece ?
16. The sum of two numbers is 120, and their difference is
20. What are the numbers ?
17. Divide $ 1800 between two persons, giving to one $ 650
more than to the other.
18. A man bought a watch and chain for $ 85, and the cost
of the watch was $ 5 more than 3 times the cost of the chain.
What was the cost of each ? *
19. In a certain election 364 votes were polled by the two
parties, and one party had 48 majority. How many votes
were cast by each party ?
20. In a certain village, containing 327 persons, there are
15 more women than men, and twice as many children as there
are men and women together. How many of each in the
village ?
21. Three men. A, B, and C, bought a mill for $ 12,000. A
paid twice as much as B, and paid 5 times as much as B.
How much did each pay ?
22. Divide a piece of cloth containing 42 yards into three
pieces, making the second piece 3 times the length of the
first, and the third piece one half of the length of the other
two pieces together.
23. Cut a cord 45 feet long into two pieces such that one
piece shall be 15 feet longer than the other.
24. Divide $8400 among A, B, and C, giving to B twice as
much as to A, and to C twice as much as to B.
25. Divide $18 among A, B, and C, giving to A twice as
much as to B, and to C twice as much as to A and B together.
26. Divide $ 1800 among three persons, giving to the second
$ 200 more than to the first, and to the third $ 200 more than
to the second.
14 ALGEBRA. [§ la
27. A is twice as old as B, and B is 15 years younger than
C, and the sum of their ages is 95 years. How old is each ?
28. Four times a certain number is 45 more than the num<
ber. What is the number ?
29. Divide $140 between two men, giving one $20 more
than the other.
30. The sum of A's and B's ages is 70 years, and A's age is
4 times B's age. What is the age of each ?
31. A school enrolls 240 pupils, and twice the mmiber of
boys equals the number of girls. How many of each are
enrolled ?
32. A man sold a horse and a buggy for $ 180, and received
twice as much for the horse as for the buggy. What was the
price of each ?
33. A mother is 3 times as old as her daughter, and the
difference of their ages is 30 years. How old is each ?
34. Divide $ 125 between A and B so that A shall receive
$ 45 more than B.
35. The sum of two numbers is 75, and their difference is
15. What are the numbers ?
36. A man owns two farms which together contain 200
acres, and the larger farm contains 42 acres more than the
smaller. How many acres in each farm ?
37. A farmer who owned a flock of sheep bought 3 times as
many sheep as he had, and then had 248 sheep. How many
sheep did he buy ?
38. A tree 90 feet long was broken by the wind, and the
part left standing was 20 feet shorter than the part broken off.
What was the length of each part ?
39. A and B are partners in business, and A's capital is
$500 less than twice B's, and their total capital is $5500.
How much capital has each ?
§ 10.] INTRODUCTION. 16
40. An estate of $ 22,050 was bequeathed to a widow and
two sons. The sons received equal shares, and the widow
twice as much as the two sons together. How much did each
receive ?
41. Divide a line 64 inches long into two parts such that the
longer shall be 8 inches less than twice the shorter.
42. A banker paid f 102 in tendollar, fivedollar, and two
dollar bills, using the same number of bills of each kind. How
many bills of each kind did he use ? How many bills in all ?
43. A school, enrolling 76 pupils, is divided into three
classes. There are twice as many pupils in the second class
as in the first, and 16 more pupils in the third class than in
the second. How many pupils in each class ?
44. A father is twice as old as his son, and the sum of their
ages less 12 years is 60 years. How old is each ?
45. A mother is 3 times as old as her daughter less 10
years, and the sum of their ages is 50 years. How old is
each?
46. A's age is twice B's, and B's age is twice C's, and the
sum of all their ages is 126 years. What is the age of each ?
47. A father and his two sons earn $140 a month, and the
father earns twice as much as the elder son, and the elder son
twice as much as the younger. How much does each earn ?
48. At an election there were three candidates. The first
received 40 votes more than the second, and 65 votes more
than the third, and the whole number of votes cast was 306.
How many votes did each candidate receive ?
16 ALGEBRA. r§ H*
CHAPTER IL
ALGEBRAIC NOTATION.
SYMBOLS REPRESENTING NUMBERS.
11. In arithmetic, numbers are represented by words and by
the Arabic numerals 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, called figures ;
and the numbers thus represented are definite known numbers.
Thus, the word eight and the figure 8 alike represent a definite
number of ones or units.
The figure 0, called zero, denotes the absence of number.
In algebra, numbers are represented by figures and by letters.
Figures in algebra, as in arithmetic, represent definite known
numbers, while letters represent any numbers, whether known
or unknown.
Figures are called arithmetical symbols of numbers; and
letters, general or algebraic symbols.
12. Known numbers, when expressed by letters, are usually
represented by the^rs^ letters of the alphabet; as a, b, c, etc.
13. Unknown numbers, that is, numbers whose values are
not given or determined, are usually represented by the final
letters of the alphabet ; as a;, y^ z.
This distinction in the use of the first and the final letters of the
alphabet is not always observed. Any letter may be used to represent
an unknown number ; and numbers may be represented by the letters of
any alphabet.
In the Roman notation, numbers are represented by the letters
/, F, X, L, C, D, My which denote definite known numbers, the same
as figures.
§ 18.] ALGEBRAIC NOTATION. 17
14. The term quxirvtity is used in algebra as synonymous
with number, A number expressed by letters, or by letters
and figures in combination, may be called an algebraic quantity
or an algebraic number. In this treatise the term number
instead of quantity ^ is generally used.
SIGNS OF OPERATION.
15. In both arithmetic and algebra the operations to be
performed with numbers are indicated by characters, called
signs.
16. The sign +, called plus, indicates that the number after
the sign is to be added to the number before it. Thus, 7 f 5,
read " 7 plus 5," denotes that the number 5 is to be added to
the number 7 ; and a\b denotes that the number repre
sented by 6 is to be added to the number represented by a, or,
more briefly, that the number b is to be added to the number a.
17. The sign — , called minus, indicates that the number
after the sign is to be subtracted from the number before it.
Thus, 7 — 5, read " 7 minus 5," denotes that the number 5 is to
be subtracted from the number 7 ; and a — b denotes that the
number represented by b is to be subtracted from the number
represented by a, or, more briefly, that the number b is to be
subtracted from the number a.
The addition of a and b is denoted by a + 6, and the sub
traction of b from a by a — 6.
1. Express the addition of x and y\ 2x and 3y ; 2 and y,
2. Express the addition of a and 2 6; 2 a and 36; a, 2b,
and 15 ; 3 a, 2 6, and 5 c; 5, 4 a, and 6 6.
3. Express the subtraction of y from a; x from 3y; 2
from X] X from 2; 3y from 5x.
18. The sign x , called the sign of multiplication, and read
"multiplied by" or "times," indicates that the two numbers
white's alo. — 2
18 ALGEBRA. [§ 19.
between wbicli it is placed are to be multiplied together.
Thus, a X b denotes that the number a is to be multiplied by
the number b, or the number b by the number a. The num
bers multiplied together are called factors; and the result
obtained by multiplying together the factors, the product.
19. When one or both of the numbers to be multiplied are
represented by letters, their multiplication may be expressed
(1) by the sign x between them, as a x & or 3 x 5 ; (2) by a
dot between the numbers, as a • 6 or 3 • 6 ; or (3), as is more
common, by writing the numbers together, os ctb ot Sb,
When numbers are expressed by figures only, their mul
tiplication is indicated by the sign x or the dot, as 12 x 5
or 12 . 5.
20. The sign f, called the sign of diyision, and read " divided
by," indicates that the number before it is to be divided by the
number after it. Thus, a * 6 denotes that the number a is to
be divided by the number b.
The division of one number by another may also be indicated
by writing the dividend over the divisor in the form of a frac
tion. Thus, the division of a by 6 may be indicated by a * 6,
or , often written "/j.
b
21. In arithmetic the operations indicated by signs can actu
ally be performed, but in algebra many operations can only be
indicated. Hence in algebra the signs +, — , x, and 5 each
may indicate both an operation and its result. Thus, a + 6
denotes the addition of a and b, and also their sum ; a — 6,
the subtraction of b from a, and also their difference ; a xb ot
a • 6, the multiplication of a and b, and also their product ; and
a f & or , the division of a by b, and also their quotient.
The expression ab indicates the product of a and b, the
sign X or • being omitted ; and usually  indicates simply the
b
§22.] ALGEBRAIC NOTATION. 19
quotient of a divided by b. In arithmetic, f denotes the
quotient of 3 by 5 as well as a fraction of a unit.
4. Express the addition of x and y; of a, y, and z.
5. Express the addition of a, 2 6, and 3c; of r, m, and n.
6. Express the subtraction of n from w ; of y from a?.
7. Express the subtraction of ab from osy] of 2 from oa;.
8. Express the multiplication ot xby y in three ways.
9. Express the multiplication of x, y, and z in three ways.
10. Write without x or • the product of a and a; of a, a:, ?/;
of 5, a,b,C', of 3, a, a;, y ; of 10, x, y, z,
11. Write without x or • the product of 3, ab, c; of 12, x, yz.
12. Express the division of aj by i/ in two ways.
13. Express the division of an by 3 a^ in two ways.
14. Let b denote the base, arid r the rate per cent, and then
express the percentage.
Suggestion. The rate per cent is the number of hundredths taken,
and hence the percentage is expressed by 6 x r, or br,
15. Let p denote the percentage, and r the rate per cent,
and express the base.
16. Let p denote the principal, r the rate per cent, and t
the time in years, and express the interest.
17. Let p denote the principal, and i the interest, and ex
press the amount.
22. The factors of a number are the numbers which, multi
plied together, will produce it. Thus, 5 and 7 are the factors
of 35 ; a and b, the factors of ab ; and 3, x, y, the factors of 3 xy.
A number that contains no integral factor except itself and 1 is called
a prime number (§ 176).
, If one factor of a product is equal to 0, the product is equal to 0, what
ever may be the finite values of the other factors.
20 ALGEBRA. [§ 23.
23. Factors expressed by figures are called mimerical factors,
and factors expressed by letters are called literal factors.
ThnSy 3 is a nuiuerical factor of Sab^ and a aad b Kteral
factors.
24. A known factor prefixed to another factor or set of
factors is called a coefficieot. Thus, in 5^7 5 is the coefficient
of a ; and in 3 ax, 3 is the coefficient of cue. a may be con
sidered the coefficient of x in ax, and 3 a in 3 ax,
25. When a coefficient is expressed by figures, it is called a
munerical coefficient ; and when it contains one or more letters,
it is called a literal coefficient. Thus, the a in ax, and the 3 a
in 3 ax, are the literal coefficients of x.
When an algebraic number has no numerical coefficient
expressed, the coefficient 1 is understood. Thus, o is the
same as la, and xy the same as 1 2^.
26. The term coefficient means cofa^ctor, and, in this sense of
the term, any factor or set of factors of a product are coeffi
cients of the other factors. Thus, in 3 ax, 3 is the coefficient
of ax, 3 a the coefficient of x, a the coefficient of 3 a?, and x the
coefficient of 3 a.
The term coefficient is, however, usually applied to the known factors,
numerical or literal, which are placed before the other factors of a
product, showing how many times they are taken.
27. A power of a number is the product ebtained by taking
the number one or more times as a factor. Thus, 25 (5 x 5)
is the second power, or square, of 5 ; 27 (3 x 3 x 3) is the third
power, or cube, of 3 ; and axaxaxa is the fourth power of a.
28. The power of a number may be indicated by writing
at the right of the number, and a little above it, a number to
denote how many times the given number is taken as a factor.
Thus, tlie third power of 5 is denoted by 5^ ; the fourth power
of a, by a^ ; and the nth power of a, by a".
§33.] ALGEBRAIC NOTATION. 21
29. The number that is taken one or more times as a factor
is called the base of the power ; and the number at its right,
denoting how many times it is taken as a factor, is called the
exponent of the power. Thus, in a^, a; is the base, and 5 the
exponent.
80. When an exponent is a figure, as in a^, it is called a
numerical exponent ; and when it is a letter, as in of", it is called
a literal exponent.
The first power of a number is the number itself ; and hence,
when a number has* no exponent expressed, the exponent 1 is
understood. Thus, 6 = 6*, and x = a;\
31 . The exponent of a number which is expressed by two or
more orders is written at the right of the righthand figure.
Thus, 25^ denotes the third power of 25 ; and 3.04*, the fourth
power of 3.04. (f )^ denotes the square of J.
32. The exponent of a literal number denotes the power of
the letter only to which it is attached. Thus, a^l^ denotes
the product of a? and 6^, and is read " a square h cube ; " and
3aV denotes the product of 3, a*, and a^, and is read "3a
fourth X cube."
33. The reciprocal of a number is unity divided by the
number. The reciprocal of 5 is  ; and of a, —
5 a
18. Write the prime factors (omitting 1) of 10; of 42 j of
11 aa;; of 6aVy; of 5ax^y^; of 21 a^h^xy^,
19. Write the numerical coefficient of 5a^; of 3.i^; of
7a:*^; of ax^i^] of 3a2f^; of f aa^^; of a^&V.
20. Write the exponent (including 1) of each factor in aWoi?)
in 5 a^a^; in ^^aWx^\ in ^a^h(^\ in ^xj^z^,
21. Express by exponents the square of oa? ; of abx ; of 2 axy,
22. Express by exponents the cube of 13 ; of aft ; of 2 abx ; of J.
22 ALGEBRA. [§ 34.
23. Express the sum of the squares of a, h, and c; the sum
of the squares of «, y, and z,
24. Express the sum of the cubes of 4, a, m, and a?; the
sum of the cubes of a, h, and 2 c.
26. Express the sum of the first three powers of x\ the
sum of the first five powers of x.
26. Assume the first number to be the minuend, and express
the difference of the squares of x and y ; of m and n ; of 3 and
a ; of 2 a and 6 ; of 3 a; and 2 y,
27. Express the product of the cubes of a, h, and c; of the
squares of x, y, and z.
28. Express in the form of a fraction the cube of a divided
by the square of h ; the square of x divided by the cube of ay.
34. The root of a number is one of the equal factors which
multiplied together will produce the number, or is the number
itself. Thus, 5 is the second or square root of 25 ; 4 is the
third or cube root of 64 ; and a is the fourth root of a\
35. The root of a number is indicated by the character y/,
called the radical sign, with a number placed above it, as y/,
called the index of the root. Thus, V9 denotes the second
or square root of 9, which is 3 ; ^125, the third or cube root
of 125, which is 5 ; and y/a, the fourth root of a.
36. When no index is expressed, the sign y/ indicates the
square root. Thus Vl6 and ^^16 each denotes the square root
of 16. The first root of a number is the number itself, and
hence V5 = 5.
The expression VST is interpreted or read " the square root
of 81 ; " ^75 + 6, " the fourth root of the sum of 75 and 6 ; "
^64 X 8, " the cube root of the product of 64 and 8 ; " and
V^, " the cube root of ^7."
29. Copy and read Va — 6 ; Va^ ; ^/a^ — b^ ; ^^214.
§41.] ALGEBRAIC NOTATION. 23
30. Copy and read Vfl? ; V^^; VaV ; V«* + 2^; v/5*.
31. What is the value, or indicated root, of s/26 ? Of ^64 ?
OfViS? Of</256?
32. What is the in dicated root of V72  8 ? Of J^4518?
Of ^/35 + 14 ? Of a/16 X 4?
33. Express the square root of 80 ; of a6; of a*— 6^; of a^+i/^.
34. Express the cube root of 18 ; of 15 x 8 ; of a+ ftj of aV.
SIGNS OF RELATION.
37. The sign =, called the sign of equality, denotes that the
numbers between which it is placed are equal (§ 5).
38. When the equality of two numbers is expressed by the
sign =, the numbers are said to be equated, and the result, is
called an equation. Thus, by equating x and a+h, we form
the equation x = a\ h,
39. The sign > or <, called the sign of inequality, denotes
that the numbers between which it is placed are unequal^ the
opening of the sign or angle being towards the greater number.
Thus, a > 6 denotes that a is greater than b, and a<b denotes
that a is less than h.
The symbol =^ denotes not equal to ; >, not greater than ; ^, not less
than. They are sometimes used.
40. The sign .*., called the sign of deduction, stands for h>ence
or therefore,
41. The sign ••• or , called the sign of continuation, stands
for the words and so on. Thus, 1  4 h 7 f 10 ••• is read
"1 + 4 + 7 + 10, and so on.''
Express the following statements by means of the proper
signs :
1. The sum of a and b is equal to c.
2. The sum of x and y is equal to the product of m and n.
3. Four times a is equal to the sum of 3 times c and twice 6.
24 ALGEBRA. [§ 42.
4. A certain number increased by 5 is equal to 3 times the
number decreased by 15.
6. The square of a number increased by 5 times the num
ber is equal to 94.
6. The sum of a and b is greater than their difference.
7. The sum of x and y is less thui their product.
SIGNS OF AGGREGATION.
42. The parenthesis ( ), the brackets [ ], the braces { ,
the vinculum , and the bar  are called signs of aggrega
tion.
43. The first four of these signs all denote that the numbers
inclosed are to be treated as one number ; that is, are to be
taken c ollect ively. Thus, 7 x (5  3), 7 x [5  3], 7 x {5  3},
and 7x5 — 3, all denote that 7 is to be multiplied by the differ
ence of 5 and 3 ; and each sign of aggregation may be removed
if 2 be substituted for 5 — 3.
44. The vinculum is used in connection with the radical
sign, as V^S + 6. The line between the numerator and the
denominator of a fraction also acts as a vinculum. Thus,
o
46. The bar denotes that the number at its right and the
number at its left are to be multiplied together. Thus,
a^ 3 xy indicates that 3xy—2y^{3y^istohe multiplied by a?,
2f
+ 31/*
Find the valuB of
1. 12 +(7 3). 5. 9x4(15 + 3).
2. 25 (13 6). 6. 23 +(8 3x2).
3. 23 (13 + 4). 7. 8(7 45)(20 + 4).
4. 17 +(7 X 5^=^). 8. 32  [(4 x 6) 12].
§46.] ALGEBRAIC NOTATION. 25
9. 96334h2}. 11. 36 15 (5x2).
10. (5+4)67^^. 12.. 8(12 7) 4(5 2).
46. When the numbers within a parenthesis, or other sign
of aggregation, are represented by letters, the operations in
dicated cannot always be performed, but the sign of aggregar
tion can be removed in the manner indicated below :
(1) a+(5 — c)=af 5 — c.
(2) (a(b + c)=a—b — c.
\a—{b — c)= a.— b + c,
(3) a x(b — c)= axb — axc = ab^ac.
(4) a^ibc)=^
It is to be specially noted that when the parenthetical
expression is preceded by the sign , as in (2) above, the
signs h and — within it must be changed (+ to — , and —
to +) when the parenthesis is removed.
The reason is obvious. The subtrahend in a — (6 + c) is & f c,
and, if b be first subtracted from a, c must be subtracted from
the result. The subtrahend in a — (5 — c) is ft — c, and, if b be
first subtracted from a, c must be added to the result, that
is, a —(6 — c)= a — 6 + c.
Remove the parentheses in each of the following expressions,
and reduce the result to its simplest form :
13. x(xy), 20. a(2 + 6) + 2(a + 6).
14. 4:X'(2x^y). 21. (a + 6)(a6).
15. x\y—(xy). 22. 5(a + 6)(3a 26).
16. a2(a*4). 23. {ab)b (ab jb^,
17. ax {ax {a), , ^24. 2(ab  b^  (2 a  b)b.
18. Sab(2abb), 25. a^2xy\f (x^^f),
19. a^b(a^bb + 4:). 26. a^y^(x^xy\y'^.
28 ALGEBRA. [§ 53.
63. A polynomial that consists of two terms, as a + 6, is
called a binomial; and one that consists of three terms, as
a' f 2 oft h &*, is called a trinomial.
An algebraic expression that consists of only one term is also called
simple ; and one that consists of two or more terms, compound,
64. Like terms are algebraic terms that have the same literal
factors. Thus, dbx^ 3 ahXy and — 8 abx are like terms. Like
terms are also called similar terms.
66. The dejg^ee of a term is the number of its literal factors.
Thus, a and 5 x are each of the first degree ; 3 a^ and 3 ab, each
of the second degree ; and 5 a^ar^, a^hx, and — 3 ca?, each of the
fourth degree.
56. A polynomial is homogeneous when all its terms are of
the same degree. Thus, the polynomial 3aj^— 6a^2^+5a^i2/^
is homogeneous, each term being of the third degree.
Copy and read the following algebraic expressions; then
give the number of terms in each, and the degree of each term :
1. 3a63a2 + 6c. 7. S ax" \ 11 a%^x ^ 5 aJ'y  9 a^,
2. 5a^>2 4.3a«2a6^ 8. 3 a^a? + 5(a2 + a^^  a^a^.
3. 4a,23a^ + 22^. ^ (^ + y)(^ V) ^  ^ f
4. a^\2xy+f, 10. 3a^~if^ + ^^.
5. 15aVGaa^h5oc^. 11. 3(a { b)  5(a  b) + S ab,
6. 4.a%^4^a'b\ 12. 2 a  ^^^i^ + 4(a + 6).
3 4
13. a^ + 4a^?/ + 6ajy + 4a;.v^ + y*.
14. x' + 2xy\y^2{x'\y^+a?f.
15. 7a2c3 + 3aV5a«c55dL^.
c
16. 3 a^x^^ ^^" ^'^  3(a'  a^.
Z 4
17. ix^y\3x^y^'{'Sxf + y*.
§58,] ALGEBRAIC NOTATION. 29
NUMERICAL VALUE.
67. The numerical value of an algebraic expression is the
number obtained by substituting for each letter therein its
numerical value, and then performing the operations indicated.
Thus, if a = 5f 6 = 4, and c = 10, the numerical value of
4a2  6c is 4 X 52  4 X 10 = 100  40 = 60.
58. In finding the numerical value of algebraic expressions,
the following facts should be kept in mind :
I. A term preceded by no sign is positive (+)•
II. The coefficient 1 is understood when no other is given.
III. A letter has the exponent 1 when no other is given.
Let a = 5, 6 = 2, c = 3, and d = 1, in each of the following
expressions, and then find their numerical value :
1. 7a — 6 6. 13. 5 ac + 6 — 3 (6 he).
2. 5 a + 5 6c. ,  /;/2 a , 6c\ ,
^* ^ ■T■ + T)■~^^•
3. a6— 6c. \5 3y
4. 3a6 + 5c. 15. 4(a + 6)(c2).
5. abc + hcac. 16. 10 (a + 6  c)  a(c  6).
6. a^^^cc^. 17. 6(a2 + 62fc)a262.
7. a2+62 + c2. Ig _ah__^hc^a±c
8. 3a6'^562c.
6hc 2 6
9! 4a6^c^ + 5c. 19 (6^ + c^ + cP)^^.
Co
10. 5 06^ + 6c  3 6cd. 20. (6 + cf  Va f 6^.
11. 4 (a + 6) +3 (a 6). 21. (a + d)'  (6 + c)l
12. (a + 6)(ac). 22. 6* + 2 62c2 f c*.
23. (a + 6)a6(c4c?) (6c)d
24. 4a626V + 4(a2h62 + 2c + c0
25. (ah26 + c)(a226d)f362c.
30 ALGEBRA. [§ 59.
POSITIVE AND NEGATIVE NUMBERS.
59. In both arithmetic and algebra the signs + and — are
used to denote respectively addition and subtraction, as already
shown (§§ 16, 17); but in algebra, as well as in the arts and
sciences, these signs are also used to denote that numbers
belong to opposite series, that is, are opposite in quality.
When the signs f and — are thus used, they are called
signs of opposition, or, better, signs of quality.
60. Numbers may be used in opposition to denote condition,
motion, direction, time, temperature, value, etc.
Thus, if north latitude is regarded as f , south latitude is
— ; if east longitude is +, west longitude is — .
If degrees of temperature above zero are +, degrees of tem
perature below zero are — .
If distance in a given direction is f , distance in the opposite
direction is — .
If profit or gain is +, loss is — ; if credits are +, debits
are — .
61. The meaning of the signs + and — , when used to denote
quality, may be illustrated by movements in opposite directions.
A\ 1 1 1 1 1 1 1 \ 1 1 1 1 1 ± 1 iB
a c
Thus, if distance from any point in the line AB to the right,
or towards B, is considered +, distance from any point in the
line to the lefi, or towards A, is considered — .
If, for example, we start at a and move 8 spaces towards B,
the distance is denoted by f 8 ; but if we begin at a or at c
and move 5 spaces towards A, the distance is denoted by — 5.
Thus, if a man starts at a given point and walks in a straight
line to the right 8 miles, and then, turning, walks back 5
§62.] ALGEBRAIC NOTATION. 31
miles, the distance walked from the starting point is denoted
by + 8 miles, and the return distance by — 5 miles, and his
distance from the starting point is expressed by + 8 + (— 5)
= +8 — 5=+ 3. He is 3 miles to the right of the starting
point.
But if, on turning back, the man should walk 12 miles to
the left, the return distance would then be denoted by — 12
miles, and his position would be expressed by + 8 f ( — 12)
= 4 8 — 12 = — 4. He would be 4 miles to the left of the
starting point.
1. A man starts at a given point and walks 4 miles to the
right, and then turns and walks 7 miles to the left. Where
is he?
2. A ship starts at the equator and sails 50° north, and then
turns and sails 35° south. What is its latitude ?
3. A ship that is 30° north latitude sails 45° south. What
is its latitude ?
4. The temperature at noon of a certain day was 60° above
zero, and in three hours it fell 25°. What was the temperature
at 3 P.M. ?
5. The temperature at 10 a.m. was 24° above zero, and at
4 P.M. it had fallen 40°. What was the temperature at 4 p.m. ?
6. The temperature at noon of a certain day was +30°,
and at midnight it was — 15°. What was the difference in
temperature ?
62. It is thus seen that in algebra the signs + and — have
two distinct uses : one to indicate an operation^ and the other
to denote the quality of numbers.
When the sign + or — is placed between two numbers or
terms, it indicates an operation; but when the sign + or —
is placed before a monomial or before the first term of a poly
nomial, it is a sign of quality. Thus, in — a6, — d^ + he, and
— a* + oft + 6*, the sign — denotes quality.
32 ALGEBRA. [§6S.
63. The sign between any two terms of a polynomial may
"be made a sign of quality by preceding it by the sign +.
Thus, a* — 2a6 + 6* may be written a^f (— 2a^>) + ( 6*);
that is, a^ — 2ab{V expresses the sum of the terms 4 a^,
2aby +6*.
64. The sign + or — before a parenthesis is a sign of oper
ation when it is preceded by a term, as in a — (6  c) ; but it
is a sign of quality when the parenthesis stands alone, as in
— (a — b), or is the first term of a polynomial, as in —(a —6)
+ ab. The first term tcithin a parenthesis has the sign + or
— as a sign of quality, as in (— a+6c). The sign + is usually
not expressed, as in a — (6 4 c).
65. A number preceded by the sign f, expressed or under
stood, is called positive ; and a number preceded by the sign
— is called negative. When no sign precedes an algebraic
number, the sign + is understood (§ oS),
66. The signs + and — are called respectively positive and
negative. When two or more algebraic numbers have each the
sign f or — , they are said to have like signs; and when one
of two algebraic numbers is preceded by + and the other by
— , they are said to have unlike signs.
67. The value of an algebraic number, considered independ
ently of its sign, is called its absolute value. Thus, + 8 and
— 8 have the same absolute value, but + 8 has a greater alge
braic value than — 8.
The algebraic value of negative numbers decreases as their
absolute value increases. Thus, — 10 < — 1, and — 5a < — a.
Every algebraic number has an absolute value. The sign
+ or — shows that this value is positive or negative.
68. The range of algebraic numbers is double that of the
numbers in arithmetic, since the latter have no sign of quality,
the signs + and — being used in arithmetic to denote opera
tions only.
§71.] ALGEBRAIC NOTATION. 83
This double range of algebraic numbers is shown by the
series :
_8 7 6 5 4 3 2 1 0 +1 +2 +3 +4 +5 +6 +7 +8...
The numbers on the right of in this series increase from left
to right (4 4 < f 5), while those on the left of decrease from
right to left (4 > 5).
LAWS OF THE SIGNS.
69. The manner in which positive and negative numbers
are added is shown by the following equations:
+ 5 + (+3) = + 5 + 3 = + 8.
+ 2a + (+a)= + 2a + a = + 3a.
+ 5 + (3) = + 53 = + 2.
+ 2a + (a)= + 2a — a = + a.
5 + (+3) =5 + 3 =2.
2 a + (+ a) = — 2 a + a = — a.
_5 + (_3) =_5_3 =__8.
a){
(2) {
(3) {
^ ^ (2a + (a)=2aa = 3a.
70. These equations may be explained by taking the double
scale of numbers, and counting as indicated by the signs.
_8 7 6 5 4 3 2 1 0 +1 +2 +3 +4 +5 +6 +7 +8
' ' I I I 1 1 1 1 I I I I I I
Thus, in (1) begin at + 5 and count 3 spaces to the right,
to + 8, thus showing that + 5 + 3 = + 8. In (2) begin at + 5
and count 3 spaces to the left, to +2, thus showing that
+ 5 — 3 = + 2. In (3) begin with — 5 and count 3 spaces to
the right, to —2, thus showing that —5 + 3= — 2. In (4)
begin with — 5 and count 3 spaces to the left, to — 8, 'thus
showing that 5 + (3)=53 = 8.
71. Since the foregoing explanations are not dependent upon
the particular numbers 5 and 3, or the relation of the numbers
WHITENS ALO. — 3
84 ALGEBRA. [§ 72.
2 a and a, we may deduce the following laws of the signs in
addition :
I. Numbers with like signs, as in (1) and (4) in § 69, are
added by finding the sum of their absolute values, and prefix^
ing the common sign to the result,
II. Two numbers with unlike signs, as in (2) and (3), are
added by finding the difference of their absolute values, andprefioo
ing the sign of the number that has the greater absolute value,
•
More than two numbers with unlike signs may be added by finding
the sum of the positive numbers and the sum of the negative numbers,
and then adding the two results, as in II.
It is thus seen that the algebraic sum of two numbers with imlike
signs is their arithmetical difference with the sign of the greater number.
72. The following equations show the manner in which
positive and negative numbers are subtracted :
(+5(+3) = +53=+2.
Xh2a{+a) = {2a — a = + a.
+ 5(3)= + 5 + 3 = + 8.
+ 2a — (— a)= + 2a + a = + 3ct.
(1)
(2) {
/3x (5(+3)=53 = 8.
^^ (2a(+a)=2aa = 3a.
a^ 5(3)=5 + 3=2.
^^ \2a(a) = 2a\a = a.
73. To explain these equations, begin in (1) with h 5 in
§ 70, and count 3 spaces to the left, to +2, thus showing that +5
—3= +2. In (2) begin with + 5, and since —(—3) indicates
direction opposite to — 3, count 3 spaces to the right, to f 8,
thus showing that + 5 — ( 3) = + 5 + 3 = + 8. In (3) begin
with — 6 and count 3 spaces to the left, to — 8, thus showing
that —5—3=— 8. In (4) begin with —5, and, since — (—3)
indicates direction opposite to — 3, count 3 spaces to the right,
to —2, thus showing that — 5 — (— 3) = — 5 + 3 = — 2.
§77.] ALGEBRAIC NOTATION. 35
74. It is thus shown that numbers with like or unlike signs
are subtracted by changing the sign of the subtraJiend, and then
adding the resulting numbers.
76. Since multiplication is the process of taking one num
ber as many times as there are units in another number, the
multiplication of positive and negative factors may be shown
as follows :
( + ax(+&) = a + aha4to6 terms = + a&.
(2) (5x(+3) = 666 = 15;
( — ax(f6)= — a — a — a to 6 terms = — ab.
Since 3 = lll and lx8 = 8xl=8,
(3) f+5x(3)=555 = 15;
1 + a X (—&) = — a — a — a • to 5 terms = — a6.
Again, since 5x(3)=(5)(5)(5)=+545+5,
(4) (5x(3) = + 5 + 5 + 5 = + 15;
( — ax(— 6)= + a + a + a+«««to& terms = + db.
76. It is seen from (1) and (4) that the product of two
factors with like signs is positive, and from (2) and (3) that
the product of two factors with unlike signs is negative; that
is, in multiplication.
Like signs give +, and unlike signs give — .
77. Since 5 x 3 = 15, 15 h 3 = 6, and 15 s 5 = 3 ; and since
a X b = abf ab\b = a, and oft s a = 6. It is thus seen that
division is the inverse of multiplication, and hence the division
of positive and negative numbers may be shown as follows ;
(1) +ax(46) = a6; .. + a6 f (+a) = + &.
(2) — a X (+ 6) = — a6 ; .*. — a6 ^ (— a) = + &•
(3) +ax(— 6)= — a6; .*. — a6 5 (+ a) = — 6.
(4) — ax(— 6)= + a6; .*. + oft ^ (— a) = — 6.
86 ALGEBRA. [§ 7&
78. it is seen from (1) and (2) in § 77 that the quotient
obtained by the division of two numbers with like signs is posi
tftWf and from (3) and (4) that the division of two numbers
with wilike Hiffns is negative; that is, in division^
Like slyns give h, and unlike signs give — .
BXBRCISBS.
Find tlie algebraic sum of
1. 15 and 12; 15 and  12;  15 and 12;  16 and  12.
2. 8 + and 6 ; 8  6 and 5; 8 and 6  5;  8 and 6  5;
8 + and  5.
8. 12,  7, and  16 ; 36,  8,  20, and 15.
4. 73,  85, + 16, + 5 ; 80,  25,  13,  16 ;  20,15,
+ 10; 25, 13, 8, +10.
5. 5a, ~2a, +3 a, 4a; 16a, —7a, —3a, —2a', 6a,
— 6a, +4a, —3a, +7a.
Make the first number given the minuend, and find the alge^
braic difference of
6. 12 and +5; 12 and —5; —12 and +6; —12 and'
— 5,5 and — 12.
7. 23 and 8; 23 and 8; 23 and 8; 23 and 8;
— 8 and + 23 ;  8 and  23.
8. 42 and 16; 42 and +16; 42 and 16; 42 and 16;
16 and 42; +16 and +42.
9. 12a and 5a; 12a and —5a; —12a and 5a', —12a
and —5a; +5 a and — 12 a.
10. 15a and —7a; 7a and 15a; —7a and 15a; —7a
and — 15 a ; + 7 a and + 15 a.
§80.] ALGEBRAIC NOTATION. 87
Multiply
11. 27 by 4; 27 by 4; 27 by 4.
12. 3a by 9; 4a by 15; 3a by 12; Sa by 13;
12a by 5; 12a by 3.
13. 36a by 3; 36a by 3; 36a? by 3;  36aj by
2; 36 by 2x.
14. 6a6 by 5; 6a6 by —5; —5a6 by 12; 6a6 by 25;
25 by 3a6; 25by4a6.
Divide
16. 60 by 12; 72 by 8; 84 by 12.
16. 15aby— 5; — 15a?by— 3; — 40ajby8.
17. 21«by3aj; 42ajby— 7aj; — 18a;by9ic.
18. 15aa; by —5; — 33aaj by 11; — 48aaj by — 16aa:;
+ 48ajby 12a;.
LAWS OF OPERATION.
79. Algebraic operations involve several fundamental prin
ciples of so wide application that they are called Laws of
Operation.
These laws may be clearly indicated by first using particular
numbers, represented by figures, and then using letters denoting
general numbers.
80. I. The Commutatiye Law.
(7 + 3 =
(7 x3 = 3x7
' la X b =
Sum. ).^3 + 7
Piquet. .
b xa
7 + 35 = 35 + 7; etc.
a\b — c = b — c{a'y etc.
7x3x5 = 3x5x7; etc.
axbxc=bxcxa; etc.
The principle thus indicated is called the Commutative Law.
It may be stated as follows :
The sum or the product of two or more numbers is the same
in whomever order tJie numbers are taken.
86 ALGEBRA. [§ 78.
78. It is seen from (1) and (2) in § 77 that the quotient
obtained by the division of two numbers with like signs is posi
tive, and from (3) and (4) that the division of two numbers
with unlike signs is negative; that is, in division,
Like signs give f , and unlike signs give — .
EXBRCISBS.
Find the algebraic simi of
1. 15 and 12, 15 and  12;  15 and 12;  15 and  12.
2. 8 46 and 5; 8 6 and 5; 8 and 6 5; 8 and 6 5;
8 + 6 and 6  5.
3. 12,  7, and  16 ; 36,  8,  20, and 15.
4. 73, 85, +16, +5; 80, 25, 13, 16; 20,15,
+ 16; 25, 13, 8, +10.
5. 5a, —2a, +3 a, 4a; 16a, —7a, —3a, —2a\ 6a,
— 5 a, + 4 a, — 3 a, + 7 a.
Make the first number given the minuend, and find the alge^
braic difference of
6. 12 and +5; 12 and —5; —12 and +5; —12 and'
— 5; 5 and —12.
7. 23 and 8; 23 and 8; 23 and 8; 23 and 8;
— 8 and + 23;  8 and  23.
8.  42 and  16 ; 42 and + 16 ;  42 and 16; 42 and 16;
16 and 42; +16 and +42.
9. 12a and 5a; 12a and —5a; —12a and 5a] —12a
and —5a; + 5 a and — 12 a.
10. 15a and —7a; 7a and 15a; —7a and 15a; —7a
and — 15 a ; + 7 a and + 15 a.
§80.] ALGEBRAIC NOTATION. 87
Multiply
11. 27 by 4; 27 by 4; 27 by 4.
12. 3a by 9; 4a by 15; 3a by 12; 6a by 13;
12a by 5; 12a by 3.
13. 36a by 3; 36a by 3; 36aj by 3;  36aj by
2; 36 by 2a?.
14. 6a6by 5; 6a& by —5; —5a6 by 12; 5a6 by— 25;
25 by 3a6; 25by4a6.
Divide
15. 60 by 12; 72 by 8; 84 by 12.
16. 15a; by 5; 15ajby3; 40 a by 8.
17. 21ajby3a;; 42a;by — 7a;; — 18a;by— 9a;.
18. 15aa; by —5; — 33aa; by 11; — 48aa; by — 16aa;;
f 48a; by 12a;.
LAWS OF OPERATION.
79. Algebraic operations involve several fundamental prin
ciples of so wide application that they are called Laws of
Operation.
These laws may be clearly indicated by first using particular
numbers, represented by figures, and then using letters denoting
general numbers.
80. I. The Commutatiye Law.
(7 + 3 =
(a + 6 =
Sum. ^7 + 3 = 3 + 7; 7 + 35 = 35 + 7; etc.
& + a; a\b — c = b — c\a', etc.
Fwduct j7x3 = 3x7; 7x3x5 = 3x5x7; etc.
^ (7 x3 = 3x7; 7
(ax6 = 6xa; a
xbxc = bxcxa'y etc.
The principle thus indicated is called the Commutative Law.
It may be stated as follows :
The sum or the product of two or more numbers is the same
in whatever order tlie numbers are taken.
8g ALGEBRA. [§ 81.
81. It follows that the terms of a polynomial or the factors
of a product may be arranged in any order, provided the signs
of the terms or factors are not changed. Thus,
^ — 2xy + f = ;ii?+f—2xy, and 3x»xy=yxa5x3.
82. II. The AsBOCiatiye Law.
Suyn,
j(12 4fi) + 7 = 12 + (r) + 7); (8 + 6)3 = 8 + (63), etc.
< (a h ^) + c = a f (6 + c) ; {a\h) — c=^a + (p — c), etc.
Pr(Hluct
((8x(J)x5=8x(6x5); 8x(6x6x4)=(8x6) x(5x4), etc.
i (axh) X c=ax (b xc) ; ax{bxcxd)=(axb)xXcxd), etc.
The principle thus indicated is called the Associatiye Law.
It may be thus stated:
The Hum or the product of three or more numbers is the same
in whatever way the numbers may be grouped.
Thifi law is the same in principle as the Commutatiye Law.
i
83. III. The Distributive Law.
Product,
[^(2 h ;j + r>)= 3 . 2 4 3 . 3 + 3 . 6 = 6 + 9 + 15;
4(4 + 3  5) = 4 . 4 + 4 . 3  4 . 5 = 16 + 12  20 ;
I a{h + c + rf) = a/> + ac + ac? ; {a + b — c)x = ax { bx — ex.
Quotient. i±i« = i> + 12; ^ + ^ + ^ = ^ + i + g.
3 3 3 a a a a
The principle thus indicated is called the Distributive Law.
It may be thus stated :
I. TJie sum of several numbers multiplied by a given number
equals the sum of the products of the several numbers multiplied
by the given number.
II. The sum of several numbers divided by a given number
equals tJie sum of the quotients of the several numbers divided by
the given number.
§85.] EQUATIONS AND PROBLEMS. 89
84. IV. The Exponent Law.
52 =55; 53 = 5.5.5. .. 52 X 5^ = 5.5. 5. 6.5 = 6* = 5*+».
d^ z=zaa\ a? = aaa, .: a? x o? = aaaaa = a* = a*"*"*.
a"'=aaa.. to m factors; a" = aaa..' to n factors. Hence
a* X a" = aaa ... to (m\n) factors = a'"^".
The principle thus indicated is called the Exponent Law. It
may be stated as follows :
JTie exponent of the product of two or more powers of a num
ber is equal to the sum of the exponents of the given powers.
Hence
The product of the several powers of a number is found by
adding their exponents. Thus, oi? x 0^ = x^'^^ = aJ^.
The above law is commonly called the Index Law ; but in this treatise
the symbol that denotes the powers of numbers is called an exponent^ and
the symbol that denotes the roots of numbers an index.
EQUATIONS AND PROBLEMS.
85. The value of x found in the solution of an equation may
be verified by substituting such value for x in the original
equation. If the two members are equal, the value of x found
is correct.
Thus, in solving the equation 5a; — 2aj + 4 = 25, the value
of X found is 7. Substituting 7 for x in the equation, we have
5 X 7  2 X 7 + 4 = 25, or 35  14 f4 = 25, or 25 = 25, and
hence 7 is the correct value of ar.
Find and verify the value of x in the following equations :
1. 7a; + 5a; 4a; = 24. 6. 10a;(3a; + 2a;)= 35.
2. 13a;6a;48a;=45. 7. 12a;(7a; 3a;)= 32.
3. 16a; 6a; 4a; = 30. 8. 2(9a;f 7a;)(3a;+5a;)=48.
4. 20a; 12a; 3a; = 40. 9. 15a;(4a;+6a;3a;)=48.
6. 8a;f(6a;7a;)=56. 10. 3(a; + 4) + 5 a; = 36.
40 ALGEBRA. [§ 85.
U. 7 a + 5(05 + 2)= 34. 16. 5(2a;6)4(a;5)=37.
12. 12a4(ah5)=36. 17. 6(a;  3)+ 3(a;f 6)= 27.
13. 9aj5(aj + l)=19. 18. 6(aj+4)(3a:f6x3)=36.
U. 13aj5(x*3)=56. 19. 5(2a;4)(6x5~2ic)=70.
15, 8(aj3)5ajh24 = 33. 20. 4(3a:5)+3(aj4) = 13.
21. A farmer raised 720 bushels of grain, consisting of
wheat, corn, and oats. He raised twice as much wheat as
corn, and 3 times as much oats as corn. How many bushels
of each grain did he raise ?
22. Divide $126 into three parts such that the second part
shall be twice the first, and the third part 3 times the second.
23. A market woman has 4 times as many dimes as quarters,
and twice as many nickels as dime^, and in all she has 52
pieces of money. How many pieces of each kind has she ?
24. In throe years a merchant made a profit of $6300. The
pnjfit the second year was twice the profit of the first year,
and tlie profit the third year was twice that* of the second
year. What was the profit each year ?
26. Three men, A, B, and C, form a partnership in business,
with a capital of $ 12,000. A furnished twice as much as B,
and C as much as A and B together. How much capital did
each furnish ?
26. The difference of two numbers is 18, and one of the
numbers is 3 times the other. What are the numbers ?
27. A has $21 more than B, and A's money is 4 times B's.
How much money has each ?
28. A father is twice as old as his elder son, and the elder
son is 3 times as old as the younger, and the sum of all their
ages is 80 years. How old is each ?
29. A newsboy, in counting his week's earnings, found that
he had twice as many dimes as quarters, and three times as
§ 85.] EQUATIONS AND PROBLEMS. 41
many nickels as dimes, and that he had in all $ 7.60. How
many pieces of money had he of each kind ?
Let X = number of quarters ; then 26 x = their value ;
2x = number of dimes, and 20 a: = their value ;
6x = number of nickels, and 30 « = their value.
Hence 25a;f 20x + 30x= 76x = 760.
30. A newsboy has $ 4.80 in quarters, dimes, and nickels,
and of each an equal number. How many pieces of money
has he? How much money in quarters, dimes, and nickels
respectively ?
31. A man and his two sons earn $162 a month, and the
man earns 3 times as much as the elder son, and the elder son
twice as much as the younger. How much does each earn ?
32. Twice A's age is 20 years more than B's age and 10
years more than C's age, and the sum of their ages is 120
years. What is the age of each?
33. A man owns two farms which together contain 180 acres
of land, and the first farm contains 20 acres more than the
second. How many acres in each farm ?
34. A farm of 640 acres was divided among a brother and
two sisters. The two sisters received an equal share, and the
brother received 40 acres less than the number of acres received
by the two sisters together. How many acres did each receive ?
36. A and B together have $180, and B has $20 less than
3 times A's money. How much money has each ?
36. A, B, and C are partners in business. A's capital is
twice B's, and C's capital is $ 500 less than 3 times B's, and
the total capital is $ 14,020. How much capital has each ?
37. Two men start at the same time from two places which
are 63 miles apart, and travel towards each other, one at the
rate of 3 miles an hour, and the other 4 miles an hour. In
how many hours will they meet ? How far will each travel ?
Suggestion. Let x = the number of hours.
42 ALGEBRA. [§85.
38. Three military companies muster together 195 men.
The second company musters 30 men more than the first, and
the third musters one half as many men as the first and second
together. How many men in each company ?
39. A lady bought 18 yards of silk and 15 yards of serge for
$ 38.25, and the silk cost twice as much per yard as the serge.
What was the cost of each per yard ?
40. A fruit dealer sold several dozens of oranges at 25 cents
a dozen, twice as many lemons at 15 cents a dozen, and twice
as many pears as lemons at 10 cents a dozen, and the bill for
all was $ 3.80. How many dozens of each kind of fruit did
he sell ?
41. A man bought a suit of clothes for $ 30. The coat <Jost
$ 5 more than the trousers, and the trousers twice as much as
the vest. What was the cost of each ?
42. A woman bought a cloak, dress, and bonnet for $35.
The dress cost $8 more than the bonnet, and the cloak $4
more than the dress. What was the cost of each ?
43. A jeweler sold three watches for $ 100. He sold the
second watch for $ 15 more than the first, and the third watch
for $5 less than the second. How much did he receive for
each watch ?
44. Two boys, Charles and Harry, had an equal amount of
money. Charles paid 75 cents for a book, and Harry 50 cents
for a knife, and then Harry had twice as much money as
Charles. How much money had each at first ?
45. A farm containing 160 acres was divided among three
heirs, A, B, and C. A received 20 acres more than B, and C
received as many acres as both A and B less 40 acres. How
many acres did each heir receive?
46. Two railroad trains start from two cities 495 miles
apart, and run towards each other on the same track, one
running 30 miles an hour and the other 25 miles an hour. In
how many hours will they meet ?
§ 88.] ADDITION. 43
CHAPTER III.
ADDITION AND SUBTRACTION.
ADDITION.
86. Addition in algebra is the process of combining two or
more given numbers into one number, called the sum.
Add
1. 3 a, 5 a, 2 a, and 7 a.
2. —2 a, —5 a, —2 a, and —7 a.
3. ab, 4a6, 3a&, and 5ab.
4. — a5, — 4 a6, — 6 a6, and —lab,
5. ^a, ^a, —6a, and —la,
6. 3 aaj, — 5 ax, — 4 a>x, and 8 oa;.
7. a, b, and c.
Suggestion. The sum of a, 6, and c is a + 6 + c.
8. a, 3 h, and — d.
9. 2 a6, 3 a6, and — 13 c.
10. 3 aj, a?, 4 1/, and — y.
87. The iirst six of the foregoing examples show that like
numbers can be combined into one term, and the last four show
that unlike numbers can only be added by connecting them with
the proper sign.
88. These examples also show that algebraic numbers may
be
I. Like numbers with the same sign,
II. Like numbers vrith unlike signs,
III. Unlike numbers, i.e., no two numbers like,
IV. JSom^ numbers like, and others unlike.
44 ALGEBRA. [§ 89.
89. These facts give four classes of examples in the addi
tion of monomials.
I. Like numbers with the same sign.
Add
II. 76c, 9&C, be, 5 be.
12. —7 be, — 9 be, —be, — 5 be.
13. — 3 xy, —xy, —10 xy, —4: Qcy, —7 xy,
14. — 12 abe, — 15 abe, — abe, — 2 ahe.
15. 3a% a% 12 a% 5a% 13 a%.
16. 12ea?, ea^, 2ea^, ISex^, 9ea^.
17. 5(a\b), — 6(ah6), 3(a + &),and— (ah&).
18. 5xy (a — b), 3xy{a — b), 4:xy(a — b), and xy(a — b).
19. Va&, 4 Va6, 6 Va6, 9 Va6, and 12 Vo^.
20. ^mn^sfo^, — 3mnV?, — 9mnV^, and wmrsfo?.
90. To add like algebraic numbers with the same sign,
Add the eoeffieients, and to their sum annex the literal part,
and prefix the common sign.
II. Like numbers with unlike signs.
21. Add 7ab, 6ab, —2ab,  5ab, Sab.
For convenience write the numbers to be added in a column,
as at the left. Add the positive numbers and the negative
7 ab numbers, and to the arithmetical difference of the coefficients
6a& of their sums annex the literal part, and prefix the sign of the
2ab greater. Thus, 7 ab { 6ab {Sab = 16ab; and  2o6  6a6
~5a6 =7a6. 16ab 7 ab =9ab.
^^^ It is unnecessary to write the numbers to be added in a column
9 ^5 when the coefficients can be readily added. If preferred, the
positive and negative monomials may be written in separate
columns.
§ 92.] ADDITION. 45
Add
22. 7 ax, —2 ax, —5 ax, 6 ox, Sax.
23. —Sac, —ac, —5a^, — 3ac, 7 ac, 12ac.
24. —9by, —by, 5by, by, Sby, 2by.
25. 5 abx, — 2 abx, — 3 abx, 10 abx, — 4 dbx.
26. Sxz^, a»*, —7x!^, 10 a»^, — as^.
27. 8(a + c), — 3(af c), — ll(a + c), 7(a + c), 4(a + c).
28. ISixYy), h(x^y), \2(x^y), 9(aj + y), 15(a:y).
29. 15 V«, Vif, — ISVi, 3VS, —5 VS.
30. 10 a Vapy, 9 a V^, — 15 a ^xy, —7a ^sfxy, 8 a V^.
91. To add like numbers with unlike signs,
Add the positive numbers and the negative numbers, and to the
arithmetical difference of their sums prefix tlie sign of the greater.
III. Unlike numbers.
92. In arithmetic the expression 7 + 5 indicates that the
two numbers are to be added, and the result (12) is their sum ;
but in algebra the expression a + 6 not only indicates that
a and. b are to be added, but it also represents their sum (§ 21).
Hence, when the algebraic numbers to be added are unlike,
their sum is expressed by connecting them with their proper
signs in the form of a polynomial. Thus, the sum of a, b, and
— cisa + b — c.
Every polynomial may be regarded as the expressed sum of its several
terms. Thus, ab'^ — hc^ f ac^ is the sum of ab^, — bc^, and ac^.
31. What is the sum of x, y, and —z?
32. Add Sab^, —4:a% and —5c.
33. . Add x^, —2xy, and ^.
34. Add 2 «*, — 3 x^y, — 4 xy^, and 5 j^.
46 ALGEBRA. [§ 93.
36. Add ax, bx, and — ex, and reduce the sum to its simplest
form.
SuooBSTioN. dx + ftas — ca5=(a + 6 — c)x.
36. Add Saoa^ and — 9axy, and reduce to simplest form.
93. To add unlike numbers,
Connect the nurnbers to he added with their signs, and reduce
the result to its simplest form.
In algebra the results of given operations are often indicated by their
proper signs.
IV. Some Numbers Like, and Others Unlike.
37. Add 15a, 36c,  7a,  3a,  106c, — 2d, and 3.
p^ In the numbers to be added, 15 a, — 7 a,
and — 3 a are like numbers, and their alge
braic sum is 5 a ; Zhc and — 10 6c are like
numbers, and their algebraic sum is — 7 &c.
The sum sought is 5 a — 7 6c — 2 d 4 3,
16 a
la 36c
3a 106c 2d + 3
5a— 7 6c — 2d + 3 written in the form of a polynomial.
Add
38. 3 a6, 5 a% 6 a6,  8 a%  a% 6 db, and 2 ab\
39. — 3aa?, — 6ax^, 1 ax, — Sax, 7ax^, — a^y^, and 4aa?.
40. 3 Vxy, — 6a^y, — 8 Vxy, 5 Vxy, and 12a^y,
41. a^, — 2a^, — xy^, Bx^y, 4:xy^, and y^.
42. Sanm^, — 7nx^, —5ana^, 4nar^, and lOarwc*.
43. 15a6, — 66, — 7a6, 106, and3c.
44. 12 a6, — 8 6c, — 5 a6, 13 6c, and — 3 a6c.
46. 2Sx^y, Uxy^, llxhf, 20xf, and 3a^.
94. To add numbers, some like and others unlike,
Add the like numbers, and connect their suvfcs and the unlike
numbers with their proper signs.
§ 94.] ADDITION. 47
Polynomials.
1. Add 8aa — 3c2^, 3ax — 7cy, 4aaj — cy — 9a&, and 9cy
4 4 oft 4 aa.
Process.
SaxZcy ^j^^ ^^^ jU^g ^^^g .^ columns, as at the left,
Sax ley g^j^^ ^^ as in § 94. The terms of a polynomial
4 ox — qf — vao ^^^^ ^yQ arranged in any order (Commutative Law,
§ 80).
ox + 9 cy 4 4 a6
16 oaj — 2 cy — 5 a6
Add
2. 5a + bc, 2a10 6 + 3c, 8a+768c, a + 26 + 6c.
3. a + 6 + c, a — 6 + c, a + b — c, a — h — c, b — a\'C,
4. 4aj + 53^ — 62, 4y — 6aj42«, x^y — z, x + 3y + 3z.
5. 5ar^2^+22^ 2/^2 ar^432^«*5y^+22;2^ 3a^f 53/*22;2
6. 302; — 4&y — 8, — 2a2; + 563^ + 6, 7 + 6by — 5az, 5by —
Saz + 4,lSSby^5az,
7. 8a2^3c2*+2, 5af2cz^ + 5, Tcs^^lOafS, 7a/ +
02^ + 1, 52c2;2_e^2/».
8. 19aV + 86V + 2A2, a*aj2  17 6y + 50%*, 12aV +
56y10cy, 76yT15aW + 3A2.
9. 5aV)c7 aJ^c + 3 oftc^, 17 aft^c  10 a6c«, 12 aH^c  11 a«6c.
10. 4a23c« + (?, 5622a2 4 0^ + ^, 2(^762 + 5^2.
11. 72/'5aj*+a6c, 2iB*+9y23 a6c, 10 aj2+2a6c, 17 a2_ 4 y2
12. px—qy\m\n, 5 qy— px\3m—Sn, 3pa5— 4gy— 4mh7i.
13. 21 0^35 f +71 z", 822;2l7aj2_43y2, 732/2_522,2_4aJJ.
14. 18 a6 + 93 5c  101 ac, 83 oc  17 a5  23 6c, 476c
7a68ac, 3 a6 4 12 oc — 32 6c.
16. 121aj2_i44y2f492;2, 25aj*642/»162;*, 9y«169«2
362*, 42aj2  26y«  17 2^, 12 y*  33 a*  142*.
48 ALGEBRA. [§ 95.
l^x'l^pxy, 10a^7f.
17. cwr" h 6ar*  3 caj, 2 oar^ + 3 6aj^ — 4 caj, 7 oaj'^ + 4 ca?  5 6a^,
ba^ — 5cx, 7 ex — 5 ax^, 3 ca? — 4 boc^.
18. aj3+2ar^f3a;+4, 6a;5ar^+3a^2, 3ar*4a^+59a?.
95. A polynomial with, two or more terms which have a
common factor may be simplified by writing the factors not
common within a parenthesis preceded or followed by the common
factor. Thus, ax + ay — 2az may be written as a(x ^y — 2z);
a^ — 3 ax { 5 bXf as (a.*^ — 3 a + 5 b)x ; and oi^ — al) + ac^, as a^ —
a(b(^.
Simplify
19. mnx^ — mnx + mny ; 3 as? — 3 oa^ — 3 axj^ — 6 ay*.
20. Qi^ — ax — bx\(xc{ab) a^b — ah — l^d + bh.
21. a^3a^2^ + 5a^2/2. 5aj2_ 102^ + 6a; + 12 2r*.
22. ax { by — ex — dy ', a^/x — a Vy + aV2z.
SUBTRACTION.
96. Subtraction is the process of taking one number from
another.
The result obtained by faking one number from another
is called their difference.
The number subtracted is called the subtrahend, and the
number from which the subtrahend is taken is called the
minuend.
The difference of two numbers is found by subtracting the second from
the first, the number first given being the minuend. Thus, the difference
of a and 6 is a — 6.
97. If either of two numbers be taken from their sum, the
result will be the other number. Thus, 7 + 2 = 9; 9 — 2 = 7,
and 9 — 7 = 2. It is thus seen that subtraction is the inverse
of addition.
§ 101.] SUBTRACTION. 49
98. If the subtrahend and difference be added, their sum
will be the minuend. It follows that the minuend may be
regarded as the sum, and the subtrahend and difference as the
numbers added. Hence
Subtraction may be defined as the process of finding a num
ber which, added to a given number, will equal another given
number.
99. In arithmetic the subtrahend is equal to or less than
the minuend; and the minuend, subtrahend, and difference
are like numbers.
In algebra the subtrahend may be numerically greater than
the minuend, as 2 — 7 = — 5 ; and the minuend and subtrahend
may be like or unlike numbers (§ 54).
100. It has been shown in § 74 that a positive number is
subtracted by changing its sign to — , and a negative number
by changing its sign to f , and then combining the resulting
numbers as in addition.
Monomials.
1. From 5 a take 2 a ; from 5 a take —2 a.
2. From —5a take 2 a ; from —5a take —2a.
3. From 12 ab take —5ab', from — 12 ab take 7db.
4. From —1 ax take — 3 ax ; from 7 ax take — 3 ax.
5. From 2 xy take 10 xy ; from —2xy take — 10 ocy.
6. From 13 aa^y tiake —5 aa^y ; from 5 aor^^ take 12 aa?]^.
7. From x take y\ from x take — y.
8. From 3 x take 5 ; from 2 x take — 3.
101. To subtract one monomial from another,
Cliange the sign of the subtrahend, and then proceed as in
addition.
With a little practice the sign of the subtrahend can be changed men
tally ; i.e., it may be considered as changed.
white's alo. — 4
6ab 7aa;2+ 3'
3a6+ 5aa;2_ 7
50 ALGEBRA. [§ 101.
Polynomials.
9. From6a6 — 7aaj2f3 take3a6f6aa^ — 7.
Process, ^^^ convenience write the subtrahend under
the minuend, as at the left. Changing the sign,
3 ab becomes — 3 aby and —Sab and 5 ab added
are 2 ab ; changing the sign, + 5 az^ becomes
2 ab" 12 ax^ + 10 6ax2, and 6ax^ and 7ax^ are I2ax^ ; chang
ing the sign, — 7 becomes + 7, and + 7 and + 3
are + 10. The difference is 2 a6  12 ax^ + 10.
Proop. The sum of 2 ab  12 ax^+ 10 and Sab+ bax^7 ia
6 a6  7 ax2 + 3.
10. From 9 icy — 3 ar^— y take — Sxy + 5 a?— 4 y, and prove
by addition.
11. From a^ — 2xy + y^ take a^ { 2 xy + 1^, and prove by
addition.
12. From a\b take a — 6, and prove by addition.
13. From 5a^ — Sab take a^ + a6, and prove by addition
14. FromlOa— 56 + 3c take 8a — 76 f 5c.
15. From 15a^  Tab + c take 12 a* + Sab + 4c.
16. From9ar^ — 53^2/ — 8 take4a^ — 7a^ + 2.
17. Yioma^ + Sa^y\Sxf + ftQkea^Sichf + Sxfy^.
18. Fromaj«3a^y + 3a^2^takea^3a^y + 3a^ + 3/^.
19. From Sa^ bab2W take ba^bab+l^ 10,
20. From 7 a.^  2 aaj2 + 5 a^aj  a^ take 5 o^  3 a^a;  a?.
21. From a {b — c\ d — e take a — 6 — c — d + e.
22. From a — b — c — d — e take a — 6 + c — d — e.
23. From ex? + qx—pv take a^ + pv f Q'^*
24. From Ga^ + 8a6 + lOft^ take Sa^ + 6a6 + 76^
25. From j^ — 4p^aj — 6p;ii? + a:^ take p^ — 12p^a? + 7?,
26. From 21 aj^  41 aar^ f 72 a^ take 12 a^  72aaj* + 54a?.
27. From aj* f 4 aj^y + ^ ^2/^ + 2^ take 8 aj^ — 6 a^y* — y.
§102.] ADDITION AND SUBTRACTION. 61
28. From Six" hf){S(a^  b^ take 2 (a^^j^ 7(a«  6^.
29. From 6 V^T^ + Va^  b^ + 4 V3 take Va* + 6*  V3.
30. From aj^ — 3 a^a^y + aj* take ^ — Saba^y + y*
31. From aj^Sa^y +3a^ — 2r* take a:' + 3ajy»3/».
102. To subtract polynomials,
Write the subtrahend under the minuend so that like terms, if
any, shall be in the same column.
Change the signs of the terms of the subtrahend, and then pro
ceed as in addition.
Exercises in Addition and Subtraction.
1 . Add 3 (a: 4 y), 5(x f y), and (x + y),
2. Add 5Vx — y, \/x — y, and — 3 Va; — y.
3. Add aix'  f), b{Q^  y^, and 2 aio?  f).
4. Add 3aj^ + Va — x, 5a^— Va — x, and 7 ar^ — (a — a*).
5. From 2a; — 3Va?y 4 2 y take a; — ^ + 2V^.
6. From 3a^ + 2aaj + 4ar^ take a^ — oaj — ar*.
7. From a^ + Sa^chxc take a;^ — (2«^c — 3 xc).
8. Add (a — b)x, (a — b)y, and (a — b)z.
9. Add 3a(5 — x), 2a(5 — a;)— 7a(5 — x), and a(5 — a;).
10. From the sum of a^ \2xy ^ y^ and a^ — y^ take
a* — 2a?y + 2^.
11. From aa^ — 6a^ take the sum of 3a«* \5bQi:^ and — 5 oa;*
4.ba?.
12. Addy*2ar»2/^46, 52^ + 3a32/^ 3a^ + 2, and2^ + 2a:2r*
+ 3a*3/* + 4aV
13. Add 3(a + &) + 5(a6), 7(a + 6)  2(a  6), and
12(a+6)4(a6).
62 ALGEBRA. [§ 103.
14. Add 6(ar^  y2)^  5(a^  2/^,  7(0^  2^, axid 4:(a?  f).
16. Add 8(a2 ^b")^ 5(0^  b% 2(a2 +>) 12(a2  6^), and
16. Add 5(a^b\ c), — 7(a + 6 + c), ll(a + & + c), and
I3(a + 6 + c).
17. From the sum of 5c + 3aV^^, — 5 aVo(^y% and aVix^y^
take the sum of aVx^y^ and 3 c — 2 a^d^y^,
18. From a^ + 3a?2t/ f Sa^ + i/^ take a^ — ^, and from the
result take Zo^y — 3a^^.
Parentheses.
103. The subtraction of one algebraic number from another
may be indicated by writing the subtrahend in a parenthesis,
and preceding it by the sign — . Thus, o^ \'if ^{x^ — 2xy ^'i^
indicates that ar^ — 2 a^ 4 1/^ is to be subtracted from ar^ + 3^
104. If a parenthesis is preceded by the sign +> the paren
thesis may be removed without changing any of the signs within
it; but, when a parenthesis is preceded by the sign — , the signs
4 and — within it must be changed when the parenthesis is
removed (§ 46). Thus, ar^+i/^— (a^ — 2a^ + y^=a^ + y^ — ar^
+ '2xy'f = 2xy.
Conversely, any number of terms of a polynomial may be
inclosed in a parenthesis preceded by the sign + without
changing the sign of any term; but, when the terms of a poly
nomial are inclosed in a parenthesis preceded by the sign — ,
the signs of all the terms inclosed must be changed. Thus,
aS _ 3a% 4 3 aty" h^ = aJ" Za^b ^ (Sab^  b^ ;
a^ 3a^b + 3ab^ b^ = a^{{S a%  Zob^ 4 6^.
Remove the parenthesis and collect like terms in
1. 5a4& (4af26).
2. a 46 — c — (2a 4 6 4 c).
3. a2 42c2(3a2 42c2ac).
§ 104.] ADDITION AND SUBTRACTION. 63
6. 5aj — (4a5 — 3aj + 5).
Suggestion. First remove the vincalum, and then the parenthesis.
It is best for the pupil to begin always with the innermost sign of
aggregation.
6. ic^'(2xy+f)(Sa^a^2xy{y^.
7. a?— (a? — Qcy— y^^y^\ a?*).
8. 12aj«[ic*(aj2^aj2+5iB^].
9. a' + 63  [ (a* + 2 a6 + 6*  a^  6^ + 2 6*] + 4 6».
10. aj* + a:8y + 3aj22/«(arV + 3aj2y2 + y4).
11. 3aaj'2 6aj2 + 26a;4(aa^36ic* + 6a13).
12. aVaa^ar*H(3aW + 2aa^ + iB<).
13. a?V(abh^iroc^(a^Pahh^^c).
14. a + & + c — (a 4 &  c)(a  6 + c)—{c — 2a + 6).
15. 2a(36 + c) + [c(a26)],
16. 5aj[2^+(4ic2^)](aj2/).
17. 3m — w— [jo— (2m — Tw)].
18. 8m— [4n — (4m — 4n — r)].
19. a(2&c)+3c(462c)(2a + 6 + 4c).
20. 3aj[4a;+(3« + 5)(2aj6)].
21. a: + y[(3aj2/)+(a?32/)aj].4[a?(2/2aj)].
22. 2a[36(4a56)](6a76).
23. (a— 6){c— [d— (a + 2>)+ c] — (a + &)}.
24. a6 — { — 6c— [a(6 — c)— 2a6— (a6)].
26. Bx — \6y—[x—{^z — ^y)\2z{6x2y — z)']\.
26. l[la(a2 + al)(a2a2_i)].
27. a6}3c[(a+2>)(a&c)]3cJ.
28. 3[(aj  a^^) + (2 0? + 3  2 «)].
^
ALGEBRA. [§ l^^
CHAPTER IV.
MULTIPUCATIOV AMD DIYISIOIT.
MrXTIPUCATION.
IW. In alr^^ra. as in arithmetic, imiltiplicatioii is the pro
iv>i> of takr.iiT ^.•nt nuuil*er as many times as there are units in
aiivther n.:ir.l»fr. The result obtained is called the product.
Tl.e ii\i:ul»er taken or multiplied is called the multiplicand,
and the iinr.iWr deiK»tiiig how many times the multiplicand is
taken is rallt^l tlie moltipUer. The multiplicand and multi
plier are factors of the pnxUiot
In iirahiplioanoTi the pnniuct is formed from the multiplicand as the
lunhipljer is formed from unity.
106. It has l>een shown in § 75 that a xh or —ax {—b)
= \iiK and that a x {— b) or — ax6= — oft; and hence,
when two faetors have h'ke signs, their product is positive; and,
when two f;utors have f ml ike signs, their product is negative;
or, nun^ briefly stated, like signs give +, and unlike signs .
It follows that the product of several negative factors is
negative when the numl)er of factors is odd (1, 3, 5, 7, etc.),
and positive when the number of factors is eveti (2, 4, 6, etc.).
anr"law::,:;("\)><(^)=«^x(c)=a^e;
or ^a V Lm 1"""^ "" (~d)=ab xcd = abcd;
iAf c ^ ^ (^) X ((f)=a6cx (d) = a6cd.
X07. Since a^ — n ^
Vv r « ^ « a'^d «* = a X a (§ 28),
» xa = axaxaxaxa=a»:
ftiul Ronorally since a" x „» _
of anil hiter in a product ~ ""'^"' '* ^''^^^^ *^^* '^ exponent
the Hcrcral factors. '* *^"*'' ^'^ *^'* ®'*'^ °f **» exponents in
TluiB, 2 a» V 3 „s v^ „. „
§ 109.] MULTIPLICATION. 55
Monomials by Monomials.
108. 1. What is the product of 5a^b^c and Scfb<?? Of
Ba^b^c and —3a^bc\
(1) ba^b^c (2) ba'^h^c Since the monomials in (1) have like
3 a^bd^ — 3 a^bc^ signs, their product is positive ; and since
15^6^ _ 15 a558c8 in (2) they have unlike signs, their prod
uct is negative (§ 106). 5 x 3 = 15 ;
a^ X a^ = a^ ; 6^ x 6 = 6' ; c x c^ = c^. The product in (1) is 16 a^b^i^,
and in (2)  15 a^b^tfi.
In multiplying two monomials, first determine the sign of
the product, and, if —, write it; then find the product of the
coeflScients, and, if not 1, write it; and lastly annex each
letter with an exponent equal to the sum of its exponents
in the factors.
Multiply
2. 3 a^xy by 5 abx. 10. — <^yV by — 11 aa^ot^t^.
3. 12 a^bxy^ hj 3 b^ca^y^ 11.  7 a% Vd^ by a»6 V.
4. — 2 6a^ by —4 oan/^. 12. a^m^a^j^ hy — bVxz^.
6. 4: a^bf hy —2 a^bcx. 13. 20 p'^g/Yt^ hy ^pqrsH.
6.  7 oft Vaj* by 2 ac^iB*. 14. ^dabc^xy'hy IBa^b^xY^
7. 9 a^ftVy by  10 c^bxy^, 15.  12 ftV^^ar* by  bc^yz".
8. — 4 oftca?* by — 7 a^ftc^on/*. 16. — 3 x^y hy —2xy hy —3 xf.
9.8 a^c^y^T^ by — 5 b^ch/^z. 17. — 4 a^a; by ^ oa^ by — aW.
109. To multiply monomials.
Multiply the numerical coefficients, and prefix the proper sign;
then annex all the letters, giving to each an exponent equal to the
Slim of its exponents in the several factors.
Since a change in the order of the factors does not change
the product (Commutative Law, § 80), for convenience arrange
the literal factors in alphabetical order.
If a letter occurs in only one of the factors, it will have the same
exponent ia the product.
56 ALGEBRA. [§ 110.
Polynomials by Monomials.
110. Since 3(4 + 6 2) = 3 x 4 + 3 x 6 3 x 2 = 24, and
generally a(b ^ c ^ d) = ab \ ac — ad (Distributive Law, § 83),
it follows that a polynomial may he multiplied by a monomial
by multiplying each term of the polynomial by the monomial,
I. Multiply 3 a' — 5 a6 + 6 ahx by — 4a*a?.
Process.
3 a* _ 5 aft j_ ft ahx Write the two factors as at the left,
__ A 2^ and multiply each term of the multipli
cand by — 4 a%.
^ 12 rt*x + 20 a^bx  24 a^hx^
Multiply
2. a — 6 — c by— 5 ac; by 3 ac.
3. x\y — z by —x\ by 3a;; by5ajy.
4. a22a6 + 6« by 4a6; by Soft*; by 3a%.
5. 3ar* — 2iB2y + 5ir/ + 2/^ by — 3ic^2^; by4a?^.
6. Sax{2ca? — ba^hy ^ca?^ \)j —<?x.
7. a^b — 6a^aj — 6a^ 4 iC* by — bab^X] by 4a^.
8. ax^ — aV — 3a^ic + c?/^ by — ax^\ by ba^oi?^.
9. a^ftV  76y a;^ by 3a6a;; by 6a6V.
10. 12/4a;2/ + 2a;2ijy _33j22^. Ij^^3^^
II. a& + ^6c — c(Z by fac; by — ^acZ.
12. ^a262 + a26ia5bya6; by ~fa%«.
13. xfa^f + la^y^fhy^xhi^hY^x^.
111. To multiply a polynomial by a monomial,
Multiply successively each term of the polynomial by the mono
mial factor, connecting the several products with the proper signs.
§ 113.] MULTIPLICATION. 67
Polynomials by Polynomials.
112. When the multiplier in arithmetic is expressed by two
or more orders, as 45, the product is found by multiplying
successively the multiplicand by the number denoted by each
figure in the multiplier, and adding the partial products. Thus,
64 X 45 = 64 X 5 + 64 X 40 = 320 + 2560 = 2880.
In like manner in algebra, when the multiplier is a poly
nomial, the product is found by multiplying the multiplicand
by each term of the multiplier, and adding the partial products,
as shown below.
1. Multiply a«2a6 + 62 by a + 36.
Process, Write the factors as at the left. Multiply
o* — 2 a6 + 62 first each term of a^ — 2abh 6* by a, giving
a \Sb «* — 2 a^b + ab^ as the first partial product ; next
multiply by + 3 &, giving Sa^b6 ab^ + 3 68 as
the second partial product, and write the like
terms in the same column ; lastly, add the two
a*  2 a^b + ab^
Sa^beab^ + Sb^
a' + 0^6 — 5 ab^ + 3 6* partial products.
2. Multiply 63a; + 7iB2 + 8aj«byl08a + iB2.
5 Sx+ 7x2+ St^
10 8x+ x2
60 30a; +70x2 + 80 a^
 40x + 24x2 _ 66x8  64x*
6x2 3^.8+ 7x* + 8x6
60  70x + 99x2 4. 21x8  67 x* + 8x6
113. In multiplying polynomials, it is convenient to arrange
the terms of both factors in such order that the like terms of
the partial products shall fall in the same column. This is
done by selecting a letter that occurs in several terms of each
factor, and arranging the terms in both factors according to
the powers of such letter. The letter thus selected is usually
the leading letter in the two factors.
68 ALGEBRA. [§ 113.
When the powers of the selected letter increase from left
to right (as in 2 above), the polynomials are said to be
arranged according to the ascending powers of the leading
letter ; and, when the powers of such letter decrease from left
to right (as the powers of a in 1 above), the polynomials are
said to be arranged according to the descending powers of the
leading letter. The polynomial a^— 2a^6a;3a6V 6V is
arranged according to the descending powers of a and the
ascending powers of b and x,
8. Multiply 3a* f 2a  5a*bl^ + 4a'^6 by a»  3a»6 f 2a.
2 a + 3 a^ + 4 a'& — 5 a*h^ Arrange both factors according to the as
2 a + a^ — 3 a^b cending powers of a, as at the left.
4. Multiply 3 2a* + 6a? jaj* by 4a? a? 4 5.
x> — 2x^ + 6a; + 3 Arrange the multiplicand according to the de
4 x' — X + 5 Bcending powers of x, as at the left.
Multiply
6. a? + 3xyhj X'2y. 11. a; — 6a by aa; — 3a*.
6. a:*2a;3by 2a;5. 12. a^  f hj x^ { f.
7. aj4y by a; + 3^. 13. Sa^{SayhjSx^'3ay.
8. 5m*2n— 1 by m*— 3n. 14. a? + 2 ay + y* by a; — y.
9. a® h a^* by a^ — a6l 15. m*n — mn' — 3 by mn* — m*n.
10. a^2ab + b^hyab. 16. ar^ 3aj + 4 by a? 6a:.
17. a? — 2a^ + 2^ by aj* + 2a^ + ^.
18. 2a? — 4a?2/5aj2^ — 2/^ by a? — 3a^~4y*.
19. 5 a?a — 3 a?a* + a^a^ — xa* by 5 a^a? — 3xa\
20. a^\b^2ab + a^b^hja^db.
21. a?3a?y + 3a^2/33y ^_^2a^ + 2/*•
22. a^xy\f + x^y[lhjx\y — l.
23. (a^\a 20) (a  4) by a^  5a + 6.
24. (f2!f + z^(y^z)hy{yhz)(jy'Z).
26. a?3a?f 5by a?a?3.
§ 114.] MULTIPLICATION. 69
114. To multiply polynomials,
Arrange the terms of each polynomial a/icording to the ascend
ing or descending powers of the same letter.
Multiply all the terms of the multiplicand by each term of the
multiplier, and add the several partial products.
NoT£. In multiplying, observe carefnlly the laws of the signs.
EZBRCISSS.
Multiply
1. oj* — a; + 1 by a? + 1.
2. lfa + a^ + a5^by loj.
3. a^42a^ + 2^2ijy ^^y^
4. a22a6f6*by a6.
6. 7?\2xy^f\yjQ^2xy + f.
6. a22a6 + 462bya2 + 2a6 + 46l
7. 6a52.3ajy + 2/2by3aj2H5aJ2^3^.
8. la + a^a^byl+aal
9. 6aj3 + 4aj2 + 3a. + 2 by 6aj*4aj*.
10. ^ — (x?y + xy^ — 'i^hjx^y.
11. a« + a*62 + a26* + 6*by a^ft*.
12. Za^\2V6abhy a^1 ah + l.
13. a* — a^h + a6* — 6* by a + 6 — 1.
14. a?\Saa?+Za^x + a^hj 7? + 2ax\a\
15. iC* — a^ + a^ — aj + 1 by a^ + a^ — a? — 1.
16. 5aj27a; + lby 7ar^ + 2ajH4.
17. aj» + 3a; + 9by a^3aj + 9.
18. lla; + ar*244a5*by4aj + aJ^ + 5.
19. 10 + 2aj + 6a^ + aJ* by 2aj6a^ + a^.
20. aJ*aj2y2*+2^by 32/*43ar/ + 3aJ*.
21. a^ 4 6^ + c^ — a6 — oc — 6c by a + 6 I c.
22. a^2al) + l^ + (?hYa^ + 2ab + h^(?.
60 ALGEBRA. [§ 115«
28. aJ* + y* + fljy* + «"y' + ay + «Vbyx — SfL
24. (aj* + a')(a? \a)hj x — a.
26. (6 + 2 c)(6  c) by (6 f c)(6  2c).
Find the product of
26. a + 6 and a — 6; 05 j y and x^y.
27. 2a + 6 and 2a — 6; 2ajh3y and 2a; — 3y.
28. ni + n and m — n ; 2 m + n and 2 m — n.
29. x + 2,x — 2, and a? — 3.
30. af^ — x + 1, x + 2, and a? — 1.
8X, OJ f a, a; — 2 a, a? + 4 a, and a; — a.
32. a* + 2 a6 + 6*, a f 6, and a — 6.
33. aj + a, ar — a, and a^ — 2 a.
34. What is the square ofa6? Of a — 6?
36. What is the third power ofajfy? Ofa; — y?
30. Find all the powers of 2 a; + c to the third.
37. Find the fourth power of a 4 6; of a — 6.
38. Find the square of a' — 2 a6 + b\
39. Find the square ofa;fy + 2;; of x — y — z.
40. What is the third power of 3a 2c? Of3a? + 2y?
41. What is the fourth power of 2 a? — 3y ?
115. An algebraic expression may be simplified by perform
ing all the multiplications and divisions indicated, removing
the parentheses, if any, and then collecting the like terms.
Simplify
42. 3 a(a f 6 H c) 2 a(b  c).
43. (a* 4 2 a6 + b^)(a  6)  3 a(ab + b^.
44. (a + b)(b + c)  6 X [6 (a  c)].
46. (a + by(a\b)(ab)(2b^'ab).
46. 2(a«a66^(a + 6)2.
§ 120.] DIVISION. 61
47. (a + l)'(al)'.
48. (a^xy{f)^2xy(oi^f).
49. (x^\xyf)^(3^xyf)\
50. Substitute a 4 2 for a; in a^ — 2 ir* f a;.
51 Substitute a — 1 for x, and 1 — a for y, in a* + 2 ajy + y*.
DIVISION.
116. Division is the inverse of multiplication (§ 77). In
multiplication two factors are given to find their product ; in
division a product and one of its factors are given to find the
other factor. Hence
117. Division is the process of finding one of two factors
when their product and the other factor are given.
The given product is the dividend, the given factor the
divisor, and the factor sought the quotient.
Since 3x4 = 12, 12^3 = 4 and 12^4 = 3; and since
a^ X 0?:= a*, of ia?z= a\ and cf ia^ = a^, it follows, that, if
a prodtict be divided by one of its factors, the quotient will be
the other fo/ctor.
Monomials by Monomials.
118. The process of dividing one monomial by another is
easily learned if it be kept in mind that it is the inverse bi the
process of multiplying two monomials.
119. In multiplying monomials, like signs give plus, and un
like signs give minus (§ 106); and hence the same law of the
signs holds true in dividing one monomial by another.
Thus, a6 5 6 = + a, and — a& ^ (— 6) = + a; and ab i{—b)
= — a, and — a6 » 6 = — a (§ 77).
120. In multiplying monomials, the product of their coeffi
cients is the coefficient of the product (§ 109) ; and hence, in
dividing one monomial by another, the coefficient of the dividend
62 ALGEBRA. [§ 121.
is divided by the coefficiervt of the divisor, and the result is the
coefficierU of the qtwtierU. Thus, the numerical coefficient of
12a'y^3aW is 12s3 = 4.
121. In multiplying monomials, the exponents of each letter
are added (§ 109) ; and hence, in dividing monomials, the expo
nent of each letter in the divisor is subtracted from the exponent
of the same letter in the dividend. Thus, c^^c^^ or^"* = a';
and 12 a«&» * 3 M* = 3 o^'ft'^ = 3 a*b.
1. 21 a^ft* is a product, and 7a*&* is one of its two factors.
What is the other factor ?
Process. 21a»&» ^7a*l^ = (21 5 7)(a6^ a2)(&8  6«)= 3a»6.
The procesB may be illustrated by resolving both monomials into their
prime factors, and then omitting the common factors ; thus,
21 a*&» = 7 X 3 X aaaaabbb, and 7 a^b^ = 7 x adbb.
.%21a6&»^7a«62==Z2L32L«««««^ = 3xaaa6 = 3a»6.
7 X aabb
122. In the division of monomials it is often convenient to
write the dividend over the divisor in the form of a fraction,
as below.
2. Divide 2Sa^i/^ by 4aY Since 28^ ( 4) = 7,
flB^ 4 jpS = x*, y* r ^ = y«,
283^2 wid 2 ? 1 = «, then 28a^«
It is not necessary to di
Proof. — 4xV X  7xV« = 28a^«. vide « by 1 ; z may simply
be written in the qnotient.
3. Divide ^Sea'^a^ (1) by 14a««y; (2) by 14 a^ajy.
^2) ^S^ = ^«'V = 4a3x^.
§ 125.] DIVISION. 68
123. It is noted that y* is omitted in the qn^otient in Ex
ample 3^ and this may need explanation.
Since K = f(3 121), and ^ = 1, it follows that / = Ij and,
since 1 may be omitted without affecting a quotient, t/^ may
also be omitted. Any letter which would appear in a quotient
with for an exponent may be omitted, since it is equai to 1.
4. Divide ^a^^f^ by — 14aVy*.
By the rule for exponents (§ 121), — = a^« = x"«. For a full dis
Hussion of negative exponents, see § 434.
5. Divide 366^2/^ by 126%; by 186c»2^.
Divide
6. _ 63 a^^'if by  7 ojy. 11. 182 m^nf by 2^ m*ny.
7. 25 m'n'a^ by 5 mn^a^, 12. 284 a%V by  71 aVc".
8. 24 a^bca^ by  12 a'bo^ 13.  1728a^3^2r» by144a^V.
9. _ 5ep^qr by  7p^r, 14.  512 a^a^fz by  64 a^fz,
10.  120 aV«« by  3 aa^. 15. 343 mn^pj^ by  49 mn^pr.
124. To divide one monomial by another,
Divide the coefficient of the dividend by the coefficient of the
divisor, prefixing the sign  if the signs are alike, and the sign
— if the signs are unlike.
Annex to this coefficient each letter of the dividend, giving it
an exponent equal to its exponent in the dividend less its ex
ponent in the divisor.
Polynomials by Monomials.
126. Since division is the inverse of multiplication, a poly
nomial is divided by a monomial by dividing each term of the
given polynomial by the monomial, observing the laws of the
64 ALGEBRA. [§ 126.
signs. Thus, (a+6— c)a=a'fa6ac, and hence (a^a6— oc)
«a = — I =a6 — c (Distributive Law, § 83).
Qt Qi Qi
1. Divide 12 aV  8 aW/ + 4 aVj/* by 4 aV.
3 aa;2  2 xy2 + ^2^8^ quotient.
If preferred, the divisor may be written under the dividend and the
quotient at the right, after the sign =. Thus,
12aBa^~8a^a:8y« + 4a^a;V^3^a_2a^2 + ay
4a2a;2 *y r «Tr
Divide
2. 3aj»y3aj*2/*43aJ2^ by 3ajy; by 3ajy.
3. 4aW8aW + 12a%34a262 by 4a*6.
4. 2 00^2/* — 4 oic'y' — 6 00^2 by 2aajy.
6. 3aa:'y — 6aV2/^ + 6a^a^ — 6aV by —3 ay.
6. 8ajV16aj«y*4a^ by 4ic2.
7. 12aV4aV + 20aV by 2a^.
8. 25 «*^ — 20 a^a; f 45 icyaj^ by — 5 oeyz,
9. 60 a^&^c^  48 a%V + 36 a%V by 12 a6c«.
10. 5a^yz — 14: a^z^ — 6 a:i^yh + 20 a^y^^ hj — aj^.
126. To divide a polynomial by a monomial.
Divide successively each term of the polynomial by the mo
nomial.
Polynomials by Polynomials.
127. The first step in the division of one polynomial by
another is to arrange the terms of each according to the ascend
ing or descending powers of a common letter, if there be such,
as illustrated below. This order of the terms is to be kept
throughout the entire process.
§ 127.]
DIVISION.
65
1. Divide 4 a%* + a«  6«  4 a*6* by a?  h\
Arrange the terms according to the descending powers of a.
a262
a43a262 + 6*
 3 0*62 + 4 ^254
" 3 a*62 _. 3 a%^
a26*  6»
0254 _ jfi
2. Divide 6 a^V — 4aJ2/* — 4iB^y + aj* + y* by m^ht^
Arrange the terms according to the descending powers of x.
— 20^.
aj*  4a;8y + 6*2^2 ixy^ 4. y4
x2_2xyHy2
x*2a;8y+ «V
x^2xy{tf^
2a^ + 4a;^2_2xy8
xV2xy» + y*
3. Divide 24a^2/2_32aJjy4.i5aj4_ga^^2/4 jy 2aj
2xy
y.
16x*  32a% + 24x2y2 _ gxy* + 2/*
16ic* Sofiy
24x8y + 24xV*
24x8yH12xV
8x8~12x2y4.6xy2y»
12x2y28x2/»
12x2y2_ 6x2/8
2x2/8 + 2/*
 2 X2/8 + 2/*
4. Divide a* + a*6* 4 ft* by a^ab + I^.
a* + a262 4. 64
g*  a86 4 q^ft^
a %  qg62 ■!. ^58
a262 _ a68 + 54
a262 _ a58 _}. 54
a^ — aft + 62 It is observed that certain
a^ + a6 + 6^ powers of a are wanting in the
dividend. We may insert for
the second and third terms, or
leave blank spaces for these
terms, as at the left.
white's alo. — 6
66 ALGEBRA. [§ 127.
Divide
6. 6aJ*96 by 3aj6.
6. jB«3a?y + 3aV3/* by a?3a?y + 3aJ2/"^.
8. af+a^y + a^ + a^^^ict/^ + i/^hja^ + xy^f.
9. 10a*27a»6 + 34a26«18a5886*by5a26a62&l
10. 36 a*6*  60 aW + 25 a^ft* by 6 a%  6 o^.
11. a? + 2xy+fhjX'\y.
12. a^ — 2ajy + 2^ by a? — y.
13. 9a2fl2a6 + 462 by 3a + 26.
14. 25a?70xy + 4:9f hj 5x^7y.
16. 163^144 by 4aj12.
16. a'* + 6' by a + b; a* — 6* by a — 6.
17. o' + lbyajll;! — a^byl — OS.
18. 27iB^82/5 by 3aj*23^.
19. 125m«4343 by 5m^ + 7.
20. m^ — n* by m h n.
21. a* + 6* by a 4 6 ; a* — 6* by a — &.
22. 9fi — t^hjx + y',sfi — i^\)ja? — y^.
23. 125aj8 + 75a^ + 15a? + l by 5a?4l.
24. 216iB«216aaJ* + 72a2aj*8a« by 6aj*2a.
25. aj*17ic2 + 16 by aj2 + 5aj + 4.
26. a^ + a^ by aj* — aa? + a*.
27. 27a^4125 by 3a^ + 6.
28. a^ + ft' by a + &; a« + 5®by a* + 6l
29. a?ifhj a?f\ 243aj«l by 3ajl.
30. 1728 m«  343 by 12 m^  7.
31. 15aj*  14a;8+ 25aj2 6aj 9 by 3a? a? 1.
32. 14a^10ar»41a?425a;415 by 7a?5a;3.
33. aj*26aj»46aj* + 6»l by aj*6a? + l.
§130.] FRACTIONAL COEFFICIENTS. 67
34. l+fhylf+y*;lfhjl+y' + i/*.
35. «* — 2^ by 05* + y* ; also by a? — r^.
36. a^ 4 a?V + y* by a* f 7?y^ + ^.
37* a^ + 1 by a» + 1.
38, 1 aj3a^aj* by l + 2aj4a^.
39, aV — 6V by oaj* — 6Y
40, a» + 3a26+3a6* + 6^ by a2 + 2a6 + 6'.
41, 8a^36ar^ + 54aj27 by 2a?3.
42 l + /2y«by 2/2 + l2t^.
43, 6a?* + 4a^9a^y»3a^ + 22/* by 2a^ + 2ajyy».
14. 24+38aj+43a^434ar»417iC*+6ic* by 6+5a?+4x*43aj».
15. af571a^+2118aj*+13824a;+1360 by a^+5»*+36x+136.
128. To divide one polynomial by another,
Arrange both dividend and divisor according to the powers of
some common letter.
Divide the first term of the dividend by the first term of the
divisor, and write the result a« the first term of the quotient.
Multiply all the terms of the divisor by the first term of the
quotient, and subtract the product from the dividend.
If there be a remaainder, consider it as a new dividend, divide
its first term by the first term of the divisor, and proceed as
before.
FRACTIONAL COEFFICIENTS.
129. An algebraic number is integral if it has no denomi
nator that contains a letter ; and hence integral numbers may
have fractional coefficients. Thus, o? — \db \ \V\^ dji integral
trinomial with fractional coefficients.
130. The fundamental processes with fractional coefficients
are the same as those with integral coefficients.
68 ALGEBRA. [§ 131.
1. Add ^a&*4.6«c4 1 and Jaft^^ft^c^.
2. Add^a»+ia6i6^ Sa^ + :^ab{ib^y ^a^^ab + ^V,
and ^ a* — a6 + f 6*.
3. From ^«— f2^ — ficy take ix'2xy ^^y,
4. From la^^aft + ^ft^ take ^a^^oft^dl
5. Multiply aj2^^^^^ by i«i.
6. Multiply ^a22a + f by ^a + i.
7. Divide ^a^ ^a^ + ^aft^ + lfts by a2^a5 + 6«.
8. Divide ^a^ ^a^ + ^x  ^ by ^aJi.
9. Multiply aj2_a;i by ^o; J.
10. Multiply a2 + a6 + 62 ijy ^^_ J
11. Divide oj^ iaj*ia^iaj2 + ^ by a^^a? J.
12. Divide iaj8 + ^aj^ by la^ + ^aj + i.
13. Divide a«  ^a% + ia&' + &^ by ^a«a6 + i6*.
14. Divide iiC* If aj^H^ by ia^ + fa? .
16. DiYide \a^\.^xf{^f hj ^af^xy + ^y'.
DETACHED COEFFICIENTS.
131. When two polynomials can be arranged according to
the ascending or descending powers of a common letter, their
product or quotient can be readily found by operating upon
the coeflBcients detached for the purpose, and then supplying
in the result the proper powers of the letters, as shown below,
1. Multiply a« + 2a2+3a4l by a* 2a + l.
Detached coefficients, 1 + 2 + 3 + 1
1 2 + 1
1+2+3+1
_2462
1+2 + 3 + 1
1+0+03+1+1
Supplying the powers of a, a^ + a* + a^ — 3 a* + a + 1.
The product is a^  3 a^ + a + 1.
§ 131.] DETACHED COEFFICIENTS. 69
2. Multiply a«43a62268bya«4a*6 + 36».
1+0+3 2
14 + 0.}. 3
1+0+3 2
»4_012 + 8
3 + + 96
14 + 311 + 8 + 96
Product, a«  4a66 + 3a*62 _ iia«&» + Saaft* + 9a6»  66».
3. Multiply a^4i«+ibya^ia:~i.
1 + i + i
1 + i + J
iJi
Product,
4. Multiply a ^ hx \ CO? \ da? by m — na5 — ra?.
a+ 6+ c+ d
m — n — r
am + 6m + cw + dm
— an — bn— en — dn
— «r — br — cr — dr
am + (6m — an) + (cm — 6n — ar) + (c2m — en — 6r) — {dn + cr) — dr
Product,
am + (6m — an)a5+ (cm — 6n — ar)x2 + (dm — en — 6r)a5' — (dn + cr)a:* — {ir«*.
6. Multiplya2 + 2a6 46^by a22a6f6l
6. Multiply l2i»43iB2 + iB3by 2a;42ar*.
7. Multiply 2iB8  6aj2 + 5aj 2 by 3a^4ajf 5.
8. Multiply a^ — aj^y' + 2^ by cc* f 7?y^ \ y*.
9. Multiply a^  3a2 + 2a  1 by itself.
10. Multiply a^  4a2 + Ha .24 by a^ f 4ah 5.
11. Multiply 4a« + 3a» + 2a + 1 by 2a8  3a2  3.
12. Multiply ^ar» + iar^ h I a? fi by 2«*iaji.
70 ALGEBRA. [§ 131,
13. Multiply aa^\ba? + cx + dhj ma^ ^nx^r.
14. Find the square ofaj* — 3fic* — 6a; + 2.
15. Find the cube of oj* — 2 a? + 1.
16. Dividea?*2a:* + 8aj3 by aj» + 2a;l.
flc*H0 ~2x« + 8a;3 ga + 2zl
2ac8 x^HSx
2x84x242x
3x2 46x3
8x« + 6x3
Process by Detached Coefficients.
1 + 02 + 83 I lH21
li21 12 + 3
21 + 8 x22x + 3, quotient
_24+2
3 + 63
3 + 63
17. Divide2aJ* + aj88a^ + 17aj12 by 2a^3a + 4.
2 + 1 8 + 17 12 123 + 4 •
23+ 4 1+23
x^ + 2 X — 3, quotient
4
12 + 17
4
6 +
8
—
6 +
9
12
—
6 +
9
12
18. Dividea:* + 6aj»4«*+24a:27 by »^42aj — a
1 + + + 64 + 2427  1+0 + 23
1 + + 23 1 + 02 + 9
02 + 94 + 24 x82x + 9, quotient
204+ 6
9 + + 1827
9 + + 1827
19. Divide a^+aj»9aj»16aj4 by iB* + 4a + 4,
20. Divide 4a!»iB* + 4aj by 2a»3«" + 2«.
§133.] SYNTHETIC DIVISION. 71
SYNTHETIC DIVISION.
132. Au examination of the process of dividing one polyno
mial by another shows that it may be abridged by observing
the following facts :
I. Only the first term of the divisor is used as a divisor in
obtaining the successive terms of the quotient.
II. The products found by multiplying the first term of the
divisor by the successive terms of the quotient are the same as
the first terms of the corresponding dividends^ aivd hence these
products may be omitted in the process,
III. Only the first term of each successive remainder is
divided to obtain the corresponding term of the quotient, and
hence the other terms of the several remainders need not be found,
IV. The signs of the terms of the several partial products
are changed in subtracting them, and this result may be
attained by changing the signs of all the terms of the divisor
before multiplying them; but, since the first term of each
partial product is omitted (see II. above), it is only necessary
to change the signs of the rem^aining terms of the divisor,
133. The manner in which these facts may be applied in
abridging the division process is shown below. For conven
ience the divisor is written at the left in a vertical column, and
the quotient below the dividend and partial products.
1. Divide a*25a« + 40a16 by a« + 5a
Full Process,
4.
a*
Full Process,
a* + 25a2 + 40a16
a* f 6 a' — 4a^
+ 5a
6a821o2 + 40o
4
4a2 + 20a16
4o2 + 20o16
a^ _ 5 a H 4, quotient.
72
ALGEBRA.
[§ 133.
a2
Abridged Process.
04 + 25a2 + 40a 16
— 6a
+ 4
6a«+ 4aa
25 a2
20a
20a + 16
a26a + 4
+ 0+0
The quotient is a^ — 5 a + 4.
Since the first term of the divisor is used only as a divisor,
its sign is not changed, and it is separated from the succeeding
terms, whose signs are changed^ by a horizontal line.
The first term of the quotient is a^ia^ = a% written below.
The first partial product, omitting a*, is (— 5a44)xa'
= 5a»+4a2.
The second term of the quotient is —5a^ia* = — 5af
written below.
The second partial product is (— 5 a 44)x(— 5a)=25a*
20 a.
The sum of the terms in the third column is + 25 a* + 4 a* —
25 a* = + 4 a*, and the third term of the quotient is + 4 a* * a*
= h 4, written below.
The third partial product is (— 5a + 4)x4 = — 20a + 16.
The sum of the fourth and fifth columns being each 0, the
division is exact, and the quotient is a* — 5 a + 4. The num
ber of O's is one less than the number of terms in the divisor.
It is well to indicate where the division ends by a vertical
line, as above.
Process by Detached Coefficients.
1
5
+ 4
1 + 025
6+ 4
25
+ 4016
20
20 + 16
Or (2) 1
6
+ 4
1+025 + 4016
6 + 2620
+ 420+16
6+ 4+ 0+
16+ 4
+ 0+0
16+ 4
Supplying the powers of a, the quotient is a^ — 6 a + 4.
In the second process (2) the partial products are written
diagonally downward to the right, each term being opposite
§ 133.]
SYNTHETIC DIVISION.
73
the term of the divisor multiplied. If preferred, this form
may be used.
In this abridged process the several partial products or divi
dends are formed by addition or synthesis, and hence the
process is called synthetic division.
It will be showD later (§ 215) that this method is of great advantage
in factoring polynomials that cannot otherwise be readily factored. It
may also be used with advantage in finding the integral roots of certain
higher equations (§ 472).
2. Divide 2aJ* + aj822ic2 + 34a? 15 hja^ + Sxd.
1
3
+ 6
2 + 122 +3416
6 + 10
+ 15
25
9+15
25+ 3 I + 0+
The quotient is 2*2 — 5 a; + 3.
3. Divide 2ar^+3aj*8a^20ar*+30a?+60 by 2aj»5a^+ia
2 + 38 20 + 30 + 60
+ 5+0 10
+ 5
+
10
+ 20
+ 040
30+ 060
1 + 4+ 6+ 010
The quotient is x^{ix — 6, with — 10a remainder.
Divide
4. aj* — ar* — 7a^4a?+6byic2_a;_5.
5. aJ*faj«9a^16a;4 by aj2 4.43.4.4.
6. aJ*42aj2aj + 2 by aj2_a; + l.
7. H2i»2i«842a^ by l« + a^.
8. aJ«4a?* + 3aj* + 2a^ + 4aj215 by aj33
10. ixfiSa^f + 3a^y^i/hyix^Sa^y\3xy^
11. aj*^aj2^aj^ by ar* + ia; + f
+ 5.
"2/».
74 ALGEBRA. [§ 134.
CHAPTER V.
SIMPLE EQUATIONS.
134. An equation is the expression of the equality of two
numbers (§ 6).
An algebraic equation is written by means of symbols ; but it may be
read, or expressed orally, by words.
135. The number that precedes the sign of equality is called
the first member of the equation, or the left side; and the num
ber that follows the sign is called the second member, or the
right side,
136. An equation that is true only for certain particular
values of the letters therein is called an equation of condition.
Thus, 3 a? 4 3 = 15, in which x equals 4 only, is an equation
of condition.
An equation which is true for all valines of the letters therein
is called an identical equation, or, briefly, an identity. Thus
the equation a{a — h) = a^ — ah is an identity, since it is true
for all values of a and h. If, for example, a^B, and & = 4,
then 5(54) =2520; thatis,5 = 5.
In an identical equation the sign =, called the sign of
identity, may be used instead of the sign =. The identity
a 4 a = 2 a is read, " a plus a is identical with 2 a."
137. An equation which, when cleared of fractions, contains
only the first power of the unknown number, is called an
equation of the first degree. Thus, oa? — a? = a6, and a +  = ,
are equations of the first degree.
§141.] SIMPLE EQUATIONS. 75
The eqnation  = ^ ~ is not of the first degree, for, when cleared of
a X
fractions, it becomes a"^ = a^ _ ab^ an equation of the second degree.
An equation of the first degree is also called a simple
equation.
138. The solution of an equation is the finding of the value
of the unknown number in it; i.e., the finding of a number
which, substituted for the unknown number, will satisfy the
equation. The value of the unknown number is called the
root of the equation.
139. The process of solving an equation involves the mak
ing of such needed changes in the two members as do not affect
their equctXUy.
140. These changes are based on the following selfevident
truths or principles, called
AXIOMS.
1. Ifeqvxds he added to equals, the sums will be equal.
2. If equals he subtracted from equals, the remainders mil be
equal.
3. If equals be multiplied by equals, the products will be equal.
4. If equals be divided by equals, the quotients will be equal.
5. Like powers of equals are equal.
6. lAke roots of equals are equal.
7. Generally, if the same changes be made in equals, the results
wiU be equal.
141. These axioms, when applied to algebraic equations,
show that
I. The same number may be added to or subtracted from both
members of an equation. Thus, if 2; = 12, a; + 4 = 12 h 4, and
(54 = 124.
76 ALGEBRA. [§ 142.
II. Both members of an equation m^y he multiplied or divided
by the same number. Thus, if a; = 12, a? x 3 = 12 x 3 ; and
if a; = 12, aJ^3 = 12^3.
III. Like powers and like roots of both members of an equor
tion are equal. Thus, if « = 4, ar^ = 16 ; if iB* = 16, V^ = Vi6.
SOLUTION OF EQUATIONS.
142. In solving an equation, it is advantageous to transfer
the terms that contain the unknown number to the first mem
ber, and the terms that contain only known numbers to the
second member. This is called transposing the terms.
1. What is the value of x in the equation a;f4 = 16?
In x{b = a?
(1) Since a? + 4 = 16, a; + 44 = 164 (Ax. 2). .. a = 16
4 = 12.
(2) Since x + b = a, x\b — b = a — b. .\ x = a — b.
It is seen that + 4 in (1), and f 6 in (2), may be transposed
to the second member if the sign f be changed to — .
2. What is the value of x in the equation a; — 4 = 12 ?
In a; — b= a?
(1) Since a;4 = 12,a?444 = 12 + 4 (Ax. 1). .. a; = 12
+ 4 = 16.
(2) Since x — b = a,x — b\b=^a\b. .*. x = a\b.
It is seen that — 4 in (1), and — 6 in (2), may be transposed
to the second member if the sign — be changed to +.
143. In like manner it may be shown that any term may
be transposed from one member of an equation to the other,
provided its sign be changed. Thus, if 5 a; — 8 = 16  3 a;, then
5aj3a; = 16 + 8. .. 2 a: = 24, and a; = 12.
144. The signs of all the terms of an equation may be
changed, since in effect this is the same as multiplying both
members by — 1 (Ax. 3).
§146.] SIMPLE EQUATIONS. 77
145. When the terms of an equation are transposed and
the unknown terms combined in one term, the value of the
unknown number is found by dividing both members by its
coefficient. Thus, if
(1) 3aj = 30, 3iB^3 = 30^3 (Ax. 4). .•.iB = 10.
/o\ J dx b b
(2) ax = b, — = — .•.« = —
a a a
(3) (a — b)x = a + b, ^ f = — '—» .. « = — ^•
"^ a — b a — b a — b
146. When the coefficient of the unknown term is a frac
tion, the value of the unknown number is found by multiplying
both members of the equation by the denominator (Ax. 3), and
dividing both resulting members by the numerator. Thus, if
(1) faj = 15, 3a?=15 x 4 = 60, and aj = 60^3 = 20.
2ab
a
(2) — aj = 2 a, oaj = 2 a&, and x = = 2 6.
It is really unnecessary to refer to Axioms 3 and 4 in the above solu
tions : for if 3a; = 30, it is evident that ac = J of 30, or 30 ^ 3 = 10 ; and
if Jaj = 15, it is evident that 3x = 16 x 4 = 60, and a; = 60 f 3 = 20.
Indeed, the relations involved are as evident as the axioms cited ; and,
besides, the pupil has applied these elementary principles throughout his
entire course in arithmetic.
The equations hitherto solved, and those to be solved in this chapter,
are integral; that is, equations no one of whose terms is fractional
(§ 129). The solution of fractional equations will be presented in
Chapter X.
Transpose the terms and find the value of x in
3. 3aj12 = 18. 9. 3a;8 = 165«.
4. 6«5 = 3aj + 10. 10. 12 aj  64 = 4(aj  5).
5. 5a? + 15 = « + 35. 11. 2(3aj4)=3a? + 13.
6. 5230?= 73 10a?. 12. 5a?(2a?+15)=274a;.
7. 12a;5 = 9aj4 13. 13. 2a? + 3 = 16 (2aj 3).
8. 3a;3 = a; + 5. 14. 5 a? 12=38 (2 a; 13).
78 ALGEBRA. [§ 146.
Find the value of x in
16. 8aj(a?12)=3aj + 16.
16. 4aj25 = 30(2iB5).
17. 6x6(x5)=7(xS)'5.
18. a? [3 +(«  3 + a?)t 2] = 7.
19. 6aj(12a?)=24(3aj4).
20. 2a?— (4a; — l)=5a?— (a?hl).
21. 6a?3(a?l)=32(5 + 2a?).
22. 5a? + 5 46(aj + 2)= 9(aj + 3).
23. 3 aj (3a?  3 4 2 a;)=2(3  ^aj).
24. (a?2)(a? + 3) = (a?5)(a? + l)+24.
26. (a?3)(a? + l) = (a?l)(a?3).
26. 7a? (2a? + 10)= 40 4 4(a? 12).
27. a? + 2(a?5)=5(13a?)3.
28. 2(a? + 3) 5(a? 7)= 3(a? 14) 1.
29. 32(a?5)=27(a?3)4.
30. 2(a? + 3)=5(a?l)(2a?l).
31. 5(a?4l)3(a?45)=2(6a?) + 2.
32. 2a?+(a — 6)=a + 6.
33. ax\b = 2a + b.
34. aa? 4 6 = a + &a?.
35. 3aa? + 6 = 2aa?— (a — 6).
36. 4:X — 2a = 6a + 2x.
37. aa? — (a — &)= 2 a + b.
38. ax — ab = b^— bx,
39. (aa? + 2)(a + l)=a(a? + 2)+5.
40. (a4a;)(aa?)4&=(2 4a)(l»).
41. oa? — 6a?=(a + 6)(a — 6).
42. a(a? — a)=6(a? — 6).
43. 2(a?&)+3(a?2 6)=2&.
44. {a'^b)x\{a—b)x^2a^b.
§150.] SIMPLE EQUATIONS. 79
147. To solve an integral equation of the first degree with
one unknown number,
Transpose all the terms containing the unknown number to
the first member, and all the other terms to the second member.
Combine the like terms in each member, and then divide both
members by the coefficient of the unknown number.
To verify the result, substitute the value of the unknown numr
ber in the original equation.
PROBLEMS.
148. A problem is a question proposed for solution.
149. The solution of a problem consists of three processes
or steps : namely,
I. The framing of algebraic expressions to denote the
different numbers in the problem, called the notation.
II. The expression of the given relations between the
known and unknown numbers in the form of an equation,
called the statement.
III. The finding of the value of the unknown number,
called the solution of the equation.
The first two of the above processes are usually called the statement.
It will be seen later that the statement of a problem may involye the
forming of two or more related equations, and its solution the finding
of the values of the imknown numbers in the several equations.
150. Any given equation may be regarded as the algebraic
statement of a problem; and the invention of problems for
which a given equation is a statement, is an interesting and
useful introduction to the inverse process. Thus, x\2x = 4t5
may be regarded as the algebraic statement of the problem,
"A's age is twice B's age, and the sum of their ages is 45
years." The papil has now had sufficient practice in the
80 ALGEBRA. [§ 150.
notation and statement of problems to make such inventions
of problems easy.
Invent problems of which the following equations are state
ments :
1. x + Sx = 48. 6. aj + 2a;f8 = 44
2. 5 a; — aj = 24. 7. a? + 24 = 6 a?.
3. 3ajfa; = 16. 8. 3aj + 4 = ajf 20.
4. 2a; a? = 18. 9. 3a?2aj = 3.
5. a; + aj + 8 = 24. 10. 2 a; f 3 a; = 400.
11. Five times a certain number less 9 equals 15 less 3
times the number. What is the number?
Let X = the number ;
then 6a;9 = 153«.
Transposing terms, 6a;43a; = 16 + 9;
combining terms, 8 a; = 24 ;
dividing by 8, x = 3.
Verification. 6x3 — 9 = 15 — 3x3;
that is, 169 = 159.
12. If 3 times a number less 6 be subtracted from 5 times
the number, the difference will equal 24 less 4 times the num
ber. What is the number ?
Let X = the number ;
then 6a;  (3a;  6) = 24  4a;.
Removing parenthesis, 5a; — 3a; + 6 = 24 — 4a;;
transposing terms, 6x — 3« + 4a; = 24 — 6;
combining terms, 6 x = 18 ;
dividing by 6, x = 3.
Verification. 6x3 (3x3 6) =24 4x3;
that is, 15  9 + 6 = 24  12, or 12 = 12.
13. Divide a line that is 25 inches long into two parts such
that the greater shall be 4 inches longer than twice the less.
§ 150.] SIMPLE EQUATIONS. 81
Let X = length of less part ;
then 2 X + 4 = length of longer part.
Hence x f 2 x + 4 = 25.
Transposing terms, xf2x = 254;
combining terms, 3x = 21;
dividing by 3, x = 7, less part ;
2 X 7 + 4 = 18, longer part.
14. A father's age is twice his son's age, and 10 years ago
his age was 3 times his son's age. What is the present age
of each ?
Let
X = son's present age ;
then
2 X = father's present age ;
2 X — 10 = father's age 10 years ago ;
X — 10 = son's age 10 years ago.
Hence
2 X  10 = 3 (X  10) = 3 X  30.
Transposing terms.
2x3x= 30 + 10;
combining terms,
 X =  20 ;
multiplying by — 1,
X = 20, son's age ;
2 X = 40, father's age.
15. The difference between two numbers is 17, and their
sum is 93. What are the numbers ?
16. What number added to 7 gives a sum equal to twice
that number increased by 1 ?
17. The sum of two numbers is 54, and their difference 10.
What are the numbers ?
18. A man is 5 years younger than his brother, and tlie
sum of their ages is 55 years. How old is each ?
19. A father is 25 years old, and his son is 5 years old. In
how many years will the father's age be twice the son's age ?
20. A father's age is 4 times the son's age, and the differ
ence of their ages is 27 years. What is the age of each ?
21. Divide $ 1000 among A, B, and C, so that A shall receive
$ 72 more than B, and C $ 100 more than A.
white's alo. — 6
82 ALGEBRA. [§ 150.
22. A man sold a horse and buggy for $ 200 ; and one third
the price of the horse was equal to one half the price of the
buggy. Find the price of each.
Suggestion. Let Sx = price of horse, and 2x = price of buggy.
23. Divide the number 72 into two parts such that 3 times
the greater may be equal to 5 times the less.
24. In a school of 143 pupils 5 times the number of boys
equals 6 times the number of girls. How many of each ?
25. I bought equal numbers of onecent, twocent, and four
cent stamps, paying $ 1.05 for all. How many did I obtain of
each kind ?
26. Divide the number 40 into three parts such that the
second may be 3 times the first, and the third double the
second.
27. At an election the number of votes cast for both can
didates was 2560, and the successful candidate had a majority
of 500 votes. How many votes did each receive ?
28. Divide $2147 between A and B so that 8 times A's
share may be equal to 11 times B's share.
29. A farmer, being asked how many sheep he had, replied
that, if he had 12 sheep more, he would have 100 less than
double the number he had. How many had he ?
30. After A had received $ 12 from B, he had $ 19 more
than B, and between them they had $75. How much had
each at first ?
31. The sum of $82.50 was paid in dollars, halfdollars,
dimes, and fivecent pieces, an equal number of each piece
being used. How many pieces of each kind ?
32. A man paid $ 1000 for a certain number of horses at
$ 60 each, 3 times as many cows at $ 30 each, and 20 times as
many sheep at $ 5 each. How many of each did he buy ?
33. A bill of $ 34 is paid in halfdollars and dimes, just 100
coins being used. How many of each were used ?
§150.] SIMPLE EQUATIONS. 88
34. A man divided $ 9 among a number of children, giving
to some a quarter each, and to twice as many a dime each.
How many children received the money ?
d5. The sum of $ 15,000 was raised among A, B, and G ; B
contributed $ 500 more than A ; and as much as A and B
together. How much did each contribute ?
36. Each of five brothiBrs is 3 years older than his next
younger brother, and the oldest is twice as old as the youngest.
What is the age of each ?
37. Divide the number 18 into two parts such that 5 times
the greater increased by 4 shall be equal to 9 times the less
diminished by 4.
38. Divide $ 4400 among A, B, and G, so that A may receive
twice B's share, and B three times G's.
39. A company of 90 persons consists of men, women, and
children. There are 4 more men than women, and 10 more
children than both men and women. How many of each in
the company ?
40. A boy engaged to carry 100 glass vessels to a certain
place on the condition that he should receive 3 cents for each
one carried safely, and pay 9 cents for each one he broke. On
settlement he received $ 2.40. How many did he break ?
41. The sum of f 5000 was divided among four persons, so
that the first and second received together $2800; the first
and third together, $ 2600 ; and the first and fourth together,
$ 2200. How much did each receive ?
42. After 34 gallons had been drawn from one of two equal
casks, and 80 gallons from the other, one cask contained 3
times as much as the other. How much did each contain at
first?
43. A brother is twice as old as his sister, and 3 years ago
he was 3 times as old as she. What is the age of each?
84 ALGEBRA. [§ 150.
44. In a brigade of 4500 men, the cavalry was 50 less than
twice the number of artillery, and the infantry was 200 more
than 8 times the cavalry. How many were there of each arm ?
45. The distance between two cities is 1083 miles. From
each a train sets out towards the other at the same hour^ one
at the rate of 35 miles an hour, and the other at the rate of 22
miles an hour. In how many hours will they meet ?
46. Divide the number 70 into two parts such that the first
plus 10 will be equal to the second multiplied by 3.
47. A and B engage in trade with equal capital. A gains
$ 1600, and B loses f 1900, and A's capital is now 8 times B's.
What was the capital of each at first ?
48. A boy has 5 more marbles in his right pocket than in
his left ; but, if he transfers 8 marbles from his left pocket to
his right, he will then have 4 times as many marbles in his
right pocket as in his left. How many marbles had he in each
pocket at first ?
49. A father gives to his four sons $ 2000, which they are
so to divide that each elder son shall receive $ 50 more than
his brother next younger. What is the share of each ?
50. A newsboy has $ 6.50 in quarters, dimes, and fivecent
pieces ; and he has 3 times as many dimes as quarters, and 5
times as many fivecent pieces as dimes. How many pieces of
each kind has he ?
51. If a steamer sails 9 miles an hour downstream, and 5
miles an hour upstream, how far can it go downstream and
return again in 14 hours ?
52. If a steamer sails downstream at the rate of a mile in
5 minutes, and upstream at the rate of a mile in 7 minutes,
how far downstream can it sail and return again in 1 hour ?
§ 154.] FORMULAS. 86
CHAPTER VI.
FORMULAS.
SPECIAL FORMS IN MULTIPLICATION AND DIVISION.
151. An algebraic equation may be the expression of a
general principle or a rule.
Thus, the equation — ^ —  — ^^— = a is the algebraic expres
sion of the principle, tJie half of the sum of two numbers added
to lialf their difference equals the greater number.
152. The algebraic expression of a general principle or rule
is called a formula.
153. The following formulas found by multiplication or
division are of special utility in abridging algebraic processes,
and also as a basis of factoring.
154. I. The square' of the sum of two numbers.
(a h by=(a + b)(a 4 &); and it is found by multiplying, that
(a + b)(a + b)= a^ f 2a6 f b\ Hence
(a\by=a^\2ab{b\ (1)
Since a and b in (1) represent any two numbers, it follows
that
The square of the sum of two numbers is the square of the first,
plus twice the product of the first multiplied by the second, plus the
square of the second.
Write by Formula (1) the square of
1. ichy. 4. a; 4 3. 7. 4ichy.
2. a + aj. 6. 5 + a?. 8. 2ax{S.
3. a? 16. 6. 3a 46. 9. 3af56.
86 ALGEBRA. [§ 155.
156. II. The square of the difference of two numbers.
(a — b)^ = (a — 6) (a — 6), and by multiplying it is found that
(a  b){a  b)= a?  2ab f W, Hence
(a  by = a*  2 a6 + b\ (2)
Since a and b represent any two numbers, it follows from
(2) that
The square of the difference of two numbers is the square of
the first f minus twice the product of the first multiplied by tlie
second, plus the square of the second.
Write by Formula (2) the square of
10. x — y. 13. 5 — X. 16. x — 5y.
11. x — a. 14. ci? — b\ 17. oaj — 4.
12. » — 3. 15. 3a — 6. 18. 4a — 36.
156. The foregoing principles are best fixed in the memory
by means of their formulas, as follows :
(a46)' = a' + 2a6 + 6l (1)
(a6)2 = a22a662. (2)
157. These two formulas may be united in one by the use
of the double sign ± , read " plus or minus," thus :
{a±by = a^±2ab^b\ (3)
If the upper sign () or the lower sign (— ) is used in the
first member of the formula, the same sign must be used in
the second member.
158. If both terms of a binomial have the same sign ia\b
or —a — b), all the terms of its square will be positive; but,
if the two terms of a binomial have unlike signs (a — b or
— a\b)j the second term of its square will be negative.
Thus, (a 4 6)' or ( a  bf = a^ f 2 a6 + 2>' J and (a bf or
(a + 6)* = a«2a6f 61
§160.] FORMULAS. 87
Write the square of
19. a\3x. 23. a* — 26. 27. —oft — 20".
20. « — 3y. 24. 2a 3b. 28. c^b — 2cx,
21. 2m 41. 25. 3a^ — 2b\ 29. aihf + 25.
22. 1— 2 m. 26. —ax + 2y, 30. 4ai^ — 2ajy.
159. III. The product of the sum and the dijfference of two
numbers.
It is found by multiplying, that
(a4.6)(a6)=a«y. (4)
Hence
The product of the sum and the difference of two numbers is
the difference of their squares.
Write the product of
31. (x '\ y)(x — y). 35. (ax — 6)(aa; f 6).
32. (aj + 4)(a?4). 36. (1 + 3a^)(l Soj^).
33. (3a; + 3)(3a;3). 37. (4a*» ~3y)(4a*a; + 33^).
34. (2a + 36)(2a36). 38. (5aW + l)(5a%2  1).
160. IV. The square of any poljrnomial.
The square of a trinomial may be readily found by Formula
(1) or (2) by including two of its terms in a parenthesis, thus
changing it to a binomial.
Thus, since a464"C=(a + 6)Hc (Associative Law, § 82),
(a 4 & 4 c)^=[(a+6)hc]*=(af6)'+2(a+6)cHc2
=^a^^2ab + ly^{2ac\'2bc + <?
= a^^b^ + <?if2ab + 2ac^2bc.
39. Find the square of a — b — c.
(o  &  c)2 = [(a  6)  c]2 = (a  6)2  2(a  6)c + C^
= a2  2 a6 + 62 _ 2 ac + 2 6c + c2
5= a^ + 62 + 0*  2 a6  2ac + 2 6c.
88 ALGEBRA. [§ 161.
40. Find the square of afftfc — d
Since a + 6}c — d=(a + &) + (c — <Z),
[(« + &) + (c  d)]2 =(a + 6)2 + 2(a + 6)(c d) + (cd)^
= o2 + 62 + c2 4 (P + 2a6 + 2ac  2ad + 2 6c  2 6d  2c(f.
161. An inspection of the foregoing equations shows that
the square of a polynomial having three terms or four terms
is made up of (1) the sum of the squares of each of its terms, and
(2) twice the product of each term multiplied in su^icession by eax^h
of the terms that follow it.
It may be shown by multiplying, that the square of any
given polynomial is the sum of the squares of its several terms
and twice the product of each term multiplied in successum b^
each of the terms that follow it.
Find the square of
41. x\y — z. 47. X \ y } z \ V,
42. X — y ^ z, 48. x — y — z — v,
43. x + y — l. 49. a — 26 — c + 2d.
44. x — y — z. 50. a + 6 — c — d.
45. 2a6 + c. 51. a^ [ 11^ — (? — d^.
46. a + 36 — c. 52. a2f622c + d.
162. V. The product of two binomials with a common term.
It may be found by multiplying, that
(1) {x\a){x\h) = a?\ax{hx\ah = a?\{a\h)x{ah\
(2) {x\a){x—h) = x^\ax—bx—ab = x^{(a—b)x—ab;
(3) (x—a)(x{b) = x^—ax{bx—ab — ar^h(— a+6)a;— a6;
(4) {x—a){x—b) = x^—ax—bx^ab = a^^{—a—b)x^ab,
163. It is seen from these formulas that the product of any
two binomials with a common term is made up of (1) the square
of the common term, (2) the product of the common term multi
plied by the algebraic sum of the two unlike terms, and (3) the
product of the unlike terms.
§ 165.] FORMULAS. 89
164. By the aid of the foregoing formulas, the product of
any two binomial factors with a common term may be written
without multiplying, care being taken to observe that the
coefficient of the second term of the product is the algebraic
sum of the two unlike terms, and that the last term of the
product is the product of the unlike terms. Thus,
(x+4:)(x  7)= a^ 3x  28, and (x4){x + 7) = a^+3x2S.
Find the product of
53. (a? + 4)(a? f 3). 61. (a?  15)(a;  9).
54. (a? — 6)(a; — 4). 62. (xa)(x — b).
65. (a;H8)(aj — 3). 63. (a  6)(a  11).
56. (x  9)(a: + 7). 64. (9a;)(9^).
67. (a; + 3)(aj  6). 66. (a;2a)(a;3).
68. (a; f 5)(aj — 13). 66. (» + 5 c)(aj  3 c).
69. (a?  10)(a;  6). 67. (a? 4 6) (a; 4).
60. (aj12)(aj + 9). 68. {x — ab)(X'\2ab),
165. VI. The product of any two binomials.
It is found by multiplying x + a by y {b that the product
of any two binomials is made ug of
I. The product of the first terms of the binomials {xy).
II. The algebraic sum of the crossproducts of the binomial
terms ; i.e., the product of the first term of each binomial mul
tiplied by the second term of the other (bx + ay).
III. The product of the second terms (a6).
Thus, (1), (2aj + 5)(x  7)= 2a^ +[2a; x ( 7) + aj x 5] 35
= 2aj29aj35.
(2), (x  b)(Sx + 26)= 3a^ +(2 bx 3bx)'2V
= 3aj*6aj26«.
90 ALGEBRA. [§ 166.
Find the product of
69. (3a?5)(»7). 77. (2a36)(a + 6).
70. (2x7){3x'{5). 78. (5a; 4y)(a;f 2y).
71. (4 m — 6)(3m — 4). 79. (2 m + 3 mn)(m — mw).
72. (3a;l)(a;3). 80. (3a 2 6) (a f 6).
73. (3aj + l)(aj2). 81. (3x + b)(x2b).
74. (aj7)(5aj12). 82. (5y  a)(2y 2a).
75. (2a? + y)(3a?22/). 83. {2 x  y)(2 x \ S y).
76. (23a6)(32a6). 84. (3a + 56)(5a  76).
DIVISION BY BINOMIAL FACTORS.
166. Since division is the inverse of multiplication, it
follows from the foregoing formulas that
a 6 aH 6
(2) a'2a& + 6'^^_^. (4. «'(^ + <')« + ^=aa
a — 6 a — 6
167. A trinomial that is the product of any two binomials
may be divided by either binomial factor by inspection.
Thus, «'3a28^^_ ^ a'3a28^
ah4 a7
Divide by inspection
1. a^2xy + f hj xy. 7. 9a*12a6+46«by 3a26.
2. a^ + 2 a% 4 «>' by a* 4 &. 8. 4aj*12«23/'+92^ by 2aj232/«.
3. ic* 6a; + 9 by a? 3. 9. a* + 5aj24 by a;3.
4. 9a*46aZ> + &* by 3a + &. 10. a' 16a + 60 by a 6.
5. 4aV+12aa;+9by2aaj+3. 11. a^ — 6aajf 6a^ by a — 3a.
6. 26»'10«2/+i/^by 6a; j/. 12. ar' + 2caj16c* by a; + 6c
§ 171.] FORMULAS. 91
168. There are several classes of binomials which are divis
ible by a binomial factor, and usually the quotient can be
written directly. The following cases are often of special
value in factoring.
169. I. Difference of two squares.
Since (a f b)(a — 6)= a* — b*,
(1) __ = a6; (2)  — ~=af ».
Hence the difference of the squares of two numbers is divish
ible by their sum or by their difference.
Divide by inspection
13. aj* — 2/* by oj — y. 18. 9aW — 6* by Sax — b.
14. aJ*2^ by aj* + 3/«. 19. 100 9a%* by 103a6.
15. aJ*  16 by a^  4. 20. 1  9 a^ by 1  3a^.
16. 9aj*9 by 3a?H3. 21. 25aVl by 5aaj*f 1.
17. 4a*96* by 2a36. 22. 16aY9 by 4ai*y3.
170. II. Sum and difference of two cubes.
It is found by dividing, that
t±^=a'^ab\b'; (1)
^tll^:=^a'^ab\b^, (2)
a — b
171. Formula (1) shows that the sum of the cubes of two
numbers is divisible by their sum, the quotient being the square
of tJie first number J minus the product of the first multiplied by
the second, plus the square of the second.
Formula (2) shows that the difference of the cubes of two
numbers is divisible by their difference, the quotient being
the square of the first number, phis the product of the first multi
plied by the second, plus the square of the second.
92 ALGEBRA. [§ 172.
Divide by inspection
23. a:^42/* by ficf y. SO. Qih^ — 7^ by xy^z.
24. 31^ — j^ hy x — y. 31. a^y^f 2;* by xy + z.
25. Sa^ + d^ by 2a + &. 32. 27 + «y by 3 + a?y.
26. 8a36« by 2a6. 33. ixfi  f by a^  f.
27. x'^lh^ by a;36. 34. aJ« + / by ar^ + j/^.
28. l8y»byl2y. 36. 1  64 ar^y^ by 1  4 iry.
29. 82/^1 by 2yl. 36. 27 aj^yS  8 by 3 a^  2.
172. III. Sum and difference of other like powers.
It is found by dividing, that
a\h
a — b
= a^a^b\ab^l^', (1)
= a^ + a^bhab^ + y'y (2)
a + 6 ' ^ ^
?LZ1^ = a* + a^ft + a«6« + a6« + 6*. (4)
a — b
It is also found by division that a^ + V, o? + 6*, and so on,
are each divisible by a + 6 ; and that cH — V, d? — l?<, and so
on, are each divisible by a — 6.
173. It will, however, be found by trial that a' + 6', a* + 6*,
a^ H 6*^, a^ + 6^ and so on, are not divisible by a + 6 or a — &,
and also that o^ — b^, a* — b^, and so on, are not divisible by
a\b.
It can be proved that, if n is any positive integer,
I. a" + b" is divisible by a f b if n is odd, and by neither
a + b nor a — b ifn is even.
II. a" — b" IS divisible bySL — bifn is odd, and by both a + b
and a — b 1/ n is even.
§ 174.] FORMULAS. 98
174. It is seen from Formulas (2) and (4) in § 172 that the
terms of the quotient are all positive when the divisor is a— 6;
and from (1) and (3), that the terms are alternately positive
and negative when the divisor is a\b. It is also seen that
the exponents of a in the quotient decrease, and those of b
increase, from left to right.
Divide by inspection
37. a^ — ^hyx + y. 44. 1 — iC* by 1 + «.
38. iB!* — ^ by a? — y. 45. 1 h «^ by 1 f a?.
39. ar^hy* by x\y. 46. 16 — aJ* by 2 — a;.
40. ar^ — y* by aj — y. 47. a:^ — 64 by a? f 2.
41. af^ — r^hyx + y. 48. aJ* — 81 by x — 3,
42. a^ — / by a — y. 49. 8a?* + l by 2ajf 1.
43. aj*l by aj1. 50. a^\27f by x + 3y.
MISCBLLAITBOUS EXBRCISBS.
Write the square of
1. x + y. 13. c h d 4 e.
2. 2 a? — y. 14. a — 6fc.
3. mhl. 15. aj + y — 2.
4. 1 — m. 16. 1 + c — d.
5. 2 3a. 17. 3a6t
6. 2xSy. 18. a26 + 1.
7. 3aj2.22^. 19. x2y\3,
8. a^^2b. 20. a + 26 + 3c.
9. 5 — be. 21. 5a — 6— c.
10. 2 aaj + y*. 22. mf7i + rha.
11. db — c. 23. m — w f r — «.
12. 7 — 2aa?. 24. ah& — c — 1.
94
ALGEBRA.
[§ 174.
Write the product of
25. (a + y)(a;y).
26. (m* — n){m^ h n).
27. (2 3/« + l)(22/«l).
28. (lh3aaj)(l — 3aic).
29. (53a;)(53a;).
30. (7af5)(7af 5).
31. (4a*f 2 6)(4a22 6).
32. (aj2 + yO(aj2y2).
33. (a^  f)(a^ ^ f).
34. (a + 7)(a — 4).
35. (» — 3a)(aj — 4 a).
36. (x f 6)(a; — c).
37. (x — m){x + n).
38. (w — 5)(m + 12).
39. (m + r)(m + s).
40. (ar*  aj + l)(aj + 1).
41. (aj« + aj+l)(al).
42. (a^'\a? + x + l){xl),
43. (a^ — aa 4 iB*)(a f x).
44. (ic* + aa + tt^(a; — a).
45. (i»*2aj + 4)(aj + 2).
46. (aj2 + 3aj + 9)(a?3).
47. Qx^ — a^ \ xf — y^(x \ y),
48. (a? + 0^ + 02^ + 2/^(0? 2^).
Write the quotient of
49.
aJ« + 2/*
ic + y
50.
a^a?
a — x
51.
o* — 2/*
«— y
52.
a^a!«
a + jc
53.
a' 64
a4
Rd
a^81
a + 3
55.
56.
57.
58.
59.
60.
aJ*9
0^ + 3
1
a*
1 +
a«
a*
1
a —
1
aj5
f
aj
y
Sa^^f
2x
+ 2/
8a»
21W
2a36
61.
62.
63.
64.
65.
66.
9aW166*
3aaj + 46*
4mV25
2m*n + 5
a25a24
a8
aj2 + 4a?32
aj + 8
a*  2 a*6^ + 6*
a26«
aJ*2a:V+2/^
x»y*
§ 179.] FACTORING. 96
CHAPTER VII.
FACTOROrG.
175. The factors of a number are the numbers which, multi
plied together, will produce the number (§ 22).
Since a x 1 = a, every number is equal to the product of
itself and 1, and hence the number itself and 1 are factors of
every number.
176. A number which has no integral factors except itself
and 1 is a prime number (§ 22).
A number which has two or more integral factors besides
itself and 1 is a composite number. Hence
177. Every composite number may be resolved into two or
more integral factors besides itself and 1.
In giving the fjwjtors of a number, the number itself and 1 are usually'
omitted. A prime factor is a prime number.
178. Factoring is the process of resolving a composite num
ber into its prime factors.
Skill in factoring is of great utility in abridging algebraic work, and
the acquisition of such skill requires much practice.
Case I.
179. Monomials.
Since 12 = 3 x 2 x 2, and a?=:a xax a, the prime factors
of 12 a^ are 3, 2, 2, a, a, a.
In like manner any monomial may be resolved into its prime
factors by factoring the coefficient, and taking the ba^e of each
Uteral factor as many times as there are units in its ea^onent (§ 29).
96 ALGEBRA. [§ 180.
Resolve into prime factors
1. 6a36'. 4. ISmViaj'. 7. 35a%V.
2. 10a*6a». 6. 49 a»cd«. 8. 81aV.
3. 9aY. 6. 121iBy. »• 100 mVs.
180. A number that is composed of tWo equal factors is a
perfect square, and one of the two equal factors of such a
number is its square root,
181. Since (Sd'f = 3a« x Sa^ = 3« x a»^« = 9a«,
V9a? = V9 X Va« = 3 x a*^« = 3 a»
Hence
The square root of a monomial is found by extracting the
square root of the coefficient, and dividing the exponent of each
letter by 2.
10. Resolve 16a*b^ into two equal factors.
Vl6a*6« = 4a26; .. 16 a*b^ = 4 a^b x 4 a^b.
Vl6 a*b^ = ±ia^b (§ 336), but in this chapter only the positive square
root is considered.
Resolve into two equal factors
11. 25 aV. 13. 121 mVr«. 15. 64a*6V.
12. Slod'f, 14. 49mVs»». 16. lUa^s^.
182. In like manner the cube root of a monomial is found
by extracting the cube root of the coeffixiient, and dividing the
exponent of each letter by 3,
Thus, ^1250^? = ^^25 x a^^b^^a^^ = 5 a%:x?.
Case II.
183. Pol3nioinial8 whose terms have a common monomial
factor.
This case presents no difficulty requiring explanation.
§ 184.] FACTORING. 97
1. What are the factors of 3aa:» H 6a%  3a«c?
It is seen by inspection that 3 a is a factor of each term^
and hence, dividing by 3 a and writing the quotient in a paren
thesis with 3 a as the other factor, we have
Sax^h^ a^by  3 a»c = 3 a (aj» f 2 a5y  a*c) ,
2. Resolve into factors a^ f a^V ■+■ ^•
««y + «*y^ + xy« = xy(x2 + xy + !^).
Resolve into factors
3. 10a% + 5a6». 8. 9 a*6a;  15 a%  3 a»6.
4. Ua^ 6Sxy^, 9. 15 a^y  10 a^ '\' 5 a^,
5. aa^  a^a^y + axf, 10. 4 a?^  12 a?y  16 a?y + 8 aV
6. a*6  a«6* + a«6». 11. 3 ary  6 «V + 9 aj^y*  12 ary.
7. cf^ + ahja^ 12. af  3 dffz + 6 a^^  aS/^,
Case III.
184. Pol3rnomial8 whose terms grouped have a common factor.
A polynomial of four terms can sometimes be so arranged
that the first two terms and the last two terms have a common
binomial factor.
Thus, oa? + ay 4 6a; ( 6y = (ax + hx) 4 (ay \by) = (a\'h)x
\{a + h)y\ and, dividing (a46)a; + (a + 6)y by a + &, we
have x\y\ and hence (a + 6) a; + (a f 6) y = (a f h){x { y).
1. Resolve into factors oar* — ay {by — ha?,
ox^ — ay \hy — hx^ = (ox* _ jx^) — (ay — 6y
= (a  6) x2  (a  6)y = (a  6)(x2  y).
The inclosing of the two terms, —ay{ by, in a parenthesis
preceded by the sign — , involves the changing of the signs
(§ 104).
2. Resolve into factors 2 aa? — 4 ay — 3 6a; f 6 6y.
2ax4ay36x + 66y = (2ax8 6x)  (^ayQhy)
= (2a36)x (2a3 6)2y=(2a36)(x2y).
white's alo. — 7
gg ALGEBRA. [§ 185.
Resolve into factors
3. a6H6y + aa;ajy. 8. s^ + y'fy + l.
4. a^ \ ax — ay — xy. 9. a' — 3 6 — a^6 + 3 a.
5. ay — ab — bx { xy. 10. aa^y + aby^ — a^a^ — a]^.
6. asy — 2 my 4 2 mn — naj. 11. 6 a' + 4 a* — 9 a — 6.
7. ax"  af \ bx"  bf. 12. 6a«a^4ay36a;»+262r'.
Case IV.
185. Trinomials which are perfect squares.
Since (a\by=a^+2ab{b'(^ 154), Va^+2a6+y =aF6;
and since (a6)2=a22a6+62(§ 155), Vc^^^2a6+^=a6.
186. It is thus seen that a trinomial is a perfect square, if,
when arranged according to the powers of some letter, its first
and last terms are perfect squares and positive, and its sec
ond term is twice the product of their square roots. Thus,
9a^\6xy {y^ is s, perfect square. Hence
The square root of a trinomial which is a perfect square is
found by connecting the square roots of the terms which are
squares with the sign of the remaining term.
Thus, ^9x'12xy + 4:f=V9x^V^=Sx2y,
187. A trinomial which is a perfect square may be resolved
into two binomial factors by extracting its square root, and mak
ing the result one of its two equal factors,
1. Resolve into factors 4 aj* — 12 aj^ + 9 y*.
V4^ = 2x; V9y* = 3y2; 2 a;  3 y2 = the square root.
Resolve into factors
2. 4ar*+4iry42/'. 6. 9 a^ — 6 xy + y',
3. a«46a5+96». 6. 4 aj*  20 ar^t/* + 25 y*.
4. a*2a«6* + &*. 7. a^2aj + l.
§ 189.] FACTORING. 99
8. Af^y + 1. 14. 440a& + 100aV.
9. 9a^'24:xy + 16f. 15. a^^^  6 ajy* + 9.
10. 25a^10xy^\^. 16. 16 + 40 oft V + 25 a% V.
11. 144 aj* 120 arV+ 25 y*. 17. 121 a*+2200a6*+ 1000061
12. 110 ay 125 ay. 18. m V  40 m W + 400 «*.
13. 25 + 30a6 + 9a%2. 19. (a + 6)2 4 (a  6)  4.
20. (aj2/)2_6(a?y)49.
21. (7n!'ny10(m^n)n + 25n\
22. a^m  a^n + b^m + 6^ — 2 a6n — 2 oftm.
Arran^ng terms, (a^w — 2 a6m + b^m) + (a^n — 2 abn + b^)
= (a2  2 aft + 62) ^ \ (^a^  2 ab + b^)n
= (a  6)2w + (a  6)2» = (a  6)2(w h n)
= (a — 6) (a — 6) (m + w).
23. a^ic* ^ 2 a^xy f a^y^ _i,2^_2 h'^xy  by.
188. Since the square of a polynomial is made up of the
sum of the squares of its several terms and twice the product
of each term multiplied in succession by all the terms that
follow it (§ 161), the square root of a polynomial which is a
perfect square may be found by connecting the square roots of
the several terms which are the squares with the proper signs,
189. The signs of the terms of the root are determined from
the several terms of the polynomial which are the products,
and this can usually be done with little difficulty by inspec
tion.
Thus, a^2bc + 2ab + <^2ac^b^
=^a2 h 6^ f c2 h 2a6  2ac 26c = (a + 6  c)*.
It is seen that each product containing c, viz., — 2ac, —2 be,
is negative, and so it is inferred that c has the sign — .
Resolve into two equal factors
24. a:^ + 2xy{2xz\f\2yz^s?.
25. a^2xy^z^ — 2yZ'\'2xz{f.
26. a*4a26 4.4a2t.462_86j.4.
100 ALGEBRA. [§ 190.
Case V.
190. Binomials expressing the difference of two squares.
Since (a \b)(ah)=a^  h^ (§ 159),
a»6*=(a + ^)(a6).
It is thus seen that the difference of two perfect squares is
equal to the product of the sum and the difference of their
square roots. Hence
191. A binomial expressing the difference of two perfect
squares is resolved into two factors by extracting the square
root of each of its terms, and then taking the sum of their roots
for one factor, and the difference of their roots for the other
factor,
1. Resolve into factors 9ajV — 4:a%
.. 90*2/2 _ 4 a252 _ (3 a.2y + 2o6)(3a;2y  2a6).
2. Resolve into factors 3 gwj* — 12 ay^.
3ax*12a2/2_3flf(a^_4y2). Vic*_a.a. V4p = 2y.
/. 3aic*12ay2_3flj(aj2 4.2 2/)(x2_2 2/).
3. Resolve into factors (m + n)^ — (m — rif,
V(m + w)2 = m + w ; V(m — n)2 = m — n.
.'. (w + w)2 — (m — w)2 =(m + w + w — n)(w + n — wi — n)
= (w + TO + w — n)(w + « — m + «)=2wx2« = 4 mn.
Resolve into factors
4. 9a*62_l6c*. 11. 2^162*.
5. ^f^.fz'. 12. 121 aV.
6. iB* — 2^. 13. iB* — 492/V.
7. 9 2/^1. 14. 14425ay.
8. 4aW36al 15. a^(a6)*.
9. 16aV492*. 16. a^{xy)\
10. 1812*. 17. 4aJ*(a46)^
§ 192.] FACTORING. 101
18. (x + yy(x^yy, 22. (a + 6)*  (a  6)*.
19. (a + 6)*(c + d)^ 23. (2a + 3)^ (3a4)«.
20. (m7i)2(m + n)*. 24. (3a + 2 6)* (2a 3 6)«.
21. (a5)«(c(f)l 26. (5 a? + 2 y)*  (a?  3 2^)1
192. Some polynomials of four or more terms can be so
written as to express the difference of two perfect squares,
and their factors can thus be readily found.
26. Resolve into factors ix^{2xy{y^ — n^.
a;2 4 2a;y + y2 _ ^ = (a;2 4. 2xy{ t/^)s^
= (x + yyz^=(x + y + z)(x {yz).
27 . Resolve into factors 4a^ — a?* + 4aj^ — 42/*.
4a2  a* + 4a;3y  43/a _ 4^2 _ (jc* _ 4x2y 4. 4y2)
= 4a2 (x2  2y)2 =(2a 4 ««  2y)(2a  x^ + 2y).
Special care must be taken to change the signs of terms when put in
a parenthesis preceded by the sign — , and also when a parenthesis pre
ceded by the sign — is removed. Thus, above, — ac*f4a;22/ — 4j/2 be
comes — («* — 4 aj2y I 4 ^) J and — (x^ — 2 y) becomes —x^\2y.
28. Resolve into factors a^ ^y^ — a^ — b^ — 2icy + 2ab,
x^ + t/^  a^  b^  2xy + 2ab = x^  2xy + 1/^ (a^  2ab + b^)
= (25  yy la  6)2 =(aj y + a b)(x  y  a + &).
Resolve into factors
29. 4a*4a*6ajy + 6^
30. 4a?*— 9ar^f 6a;l.
31. a^b^ + c'2abc25aV.
32. a^4:cx?f2ab + b\
33. l2a? + aj216a%*.
34. a2 + 46*4a64ar*92/^12a:2/.
35. a^52c2 + 2a + 26c + l.
36. a?{y^ — 2xy — 2 mn — m^ — n*.
37. i)^ + f8i^z^ + 2xy + 28Z.
88. a^2ac62_^2_2M + c2.
102 ALGEBRA. [§ 193.
193. A polynomial of the general form a*  a%^ + 6* may
be written as the difference of two squares by adding to it a^^,
and then subtracting a%* (the number added) from the result.
Thus, a* + a%^ + 6* = a* + a%* 4 2>* f a^ft^ _ a^ft*
= (a^ f by  a^h^= (a" 4 &' + a6)(a* + 6^  a6).
39. Resolve into factors a^ + a^ + 1.
= (a^ + 2x2+l)x2=(a;2+l)2a;a
40. Resolve into factors 9 a* — 3 a%* + 6*.
9a*  3a2&2 + ft* = 9a*  3a262 + 6* + 9a2&2  9a26»
= 9a* + 6a262 + 54 _ 9^2^2 =(3^2 + 52)2 _ 902^2
= (3a2 + 3a& + 62)(3a2  3a6 + 62).
Resolve into factors
41. aj* f ar^i/2 __ y4 44 ot^ + a^ft^ + ft*.
42. aj* + 2a^2^ + 92^*. 45. a:^ + 2 a^ + 9 3/*.
43. m*8mW + 4w*. 46. 4 a;*  16 a^^ + 9 2^.
194. In like manner any polynomial that can be made a
perfect square by adding to it a number, can be factored by
adding such number, and subtracting it from the result, and
then proceeding as in § 192.
Thus, a* 4 4 6*= (a* + 4 aV{4: b") A:aV= (a^{'2 b^ 4 d'b^
= {a^{2b^ + 2ab)(a^ + 2b^2ab).
47. Resolve into factors a^ 4 41/*.
a;8 + 4 y* = a;8 + 4 o^y + 4 y*  4 x^y^
= (rc* 4 2y2)2 _ 4a;V =(«* + 2y2 •^2xhf)^a^ + 2y2 _ 2a;2y).
195. If the number added and subtracted be not a perfect
square, the resulting factors will contain a radical, or indicated
root (§ 370).
Thus, a^ ^ ab + V^ :=a^ { ab + b^ f db — ab
= (a { by — db ={a i b 4 Va^)(a 4 & — Vab).
§ 199.] FACTORING. 108
Kesolve into factors
48. a?* + 4. 50. a?\ix, 52. a* + 3 a'6* + 4 6*.
49. 64 + 2^. 61. aj* + 2icy. 63. a*6a%* + 6*.
Case VI.
196. Trinomials having^ binomial factors with a common term.
Since by § 162 (a? + d){x + b) = a^ + (a { b)x + ab,
conversely ic* + (a + b)x \ a^ = (x { a)(x + b).
Hence a trinomial of the form a^ + mx \ p may be resolved
into two binomial factors, if its third term is the product of two
fcLctors whose algebraic sum is the coefficient of the second term,
197. There are two cases :
I. When the final term is positive.
II. When the final term is negative,
198. I. When the final term is positive.
By § 163, we have, conversely,
a^ + (« H b)x + aZ> = (a? + a)(aj h 6) ;
«* — (a + b)x + ab ={x — d)(x — b),
199. It is seen from these formulas, that, if the third term
of the given trinomial is positive, the second terms of its
binomial factors will have the same sign as the middle term of
the trinomial ; and (2) that the sum of the second terms of the
binomial factors is the coefficient of the middle term of the
trinomial.
1. Resolve into factors aj* + 7ic + 12.
12=3x4;3 + 4 = 7.
.'. x^+7x+ 12 =(« + 3)(a; + 4).
2. Resolve into factors aj^ — 12 a; + 35.
35 =  7 X ( 5) ; _ 7 + ( 5) =  12.
.*. »2 _ i2x + 35 =(x  7)(x  5).
104 ALGEBRA. [§ 200.
Resolve into factors
3. aj* 4 7a; 4 10. 6. 3^4.93.^20.
4. ic«13aj + 22. 7. JC* + 15a; + 66.
5. a« 14a? 4 45. 8. aj^  11 a? f 28.
200. II. When the final term is negative.
By § 162, conversely, we have
a;* f(a — b)x — ab=(x\ a)(x — b) ;
»* + (— a + b)x — ah=(x — a)(x 4 6).
201. It is seen from these equations that when the final
term of a trinomial is negative, the second terms of its two
binomial factors have unlike signs, and (2) that the algebraic
sum of the second terms of the binomial factors is the coefficient
of the middle term of the trinomial.
9. Resolve into factors a;^ + 3 a? — 28.
28=7 x(4); 74=3.
.. a;2 + 3x  28 =(a + 7)(x  3).
The third term, — 28, also equals — 7x4; but — 7 f 4 = — 3, whereas
the coefficient of the second term of the trinomial is + 3.
10. Resolve into factors ar^ — 4 a; — 45.
46 = 9x6; 9 + 5= 4.
/. x2  4x  46 =(x  9)(a + 6).
It is seen from Examples 9 and 10 that the sign of the
middle term of the trinomial is the sign of the numerically
greater second term of the binomial factors.
Resolve into factors
11. ar^ + 5aj24. 15. a^ + 10a?56.
12. a^7a?60. 16. ar^3aj~70.
13. a24aj — 45. 17. a^ — X'72,
14. a^H 2a; 63. 18. d^^x42.
202.] FACTORING. 106
202. It is thus seen that a trinomial whose binomial factors
have only one common term may be resolved into its factors
(1) by taking the sqvxire root of the first term for the first (yr comr
mon term of the two factorSy and then (2) finding for their second
terms two numbers such thcU their product is the final term of the
trinomial, and their algebraic sum the coefficient of its middle
term.
Resolve into factors
19. a^ + 8a; + 16. 38. z^'Z272.
20. a^H 14a; + 40. 39. a^ + 32 a; f 176.
21. a;2 + 16a;463. 40. a?fl0aJ76.
22. a^ + 18aj + 72. 41. aj*  40 a? f 400.
23. 2^ + 20yf96. 42. aj*faj166.
24. a^15aa; + 64al 43. f^Vly{m.
25. aj28aaj + 16a*. 44. aj*  23 a; + 76.
26. aj216aj + 66. 45. aj215a;54.
27. aj*17aj + 62. 46. f^lly42,
28. 3^272^ + 140. 47. t^7y170.
29. a^ + 3a;28. 48. f + 24.y + l^^.
30. aj« + 7aj18. 49. a^ + 17 aj« + 66.
31. a^ + 6aj66. 60. a^15a:8 + 56.
32. a:«Ha:132. 51. aj8 + 2a^99.
33. 2r^ + y182. 52. f + lfS.
34. xV9a^22. 53. 2^«»  19 y' f 48.
35. a^3/*  5 a^  104. 54. 7? + (p ^ c) x \ be,
36. a* — 7aa; — 60al 55. oi? \ (a—c)x — ac.
87. aj«2a:99. 56. y"  (a? ^l^f ^a^b
2
106 ALGEBRA. [§ 203.
203. The required factors of the final term can usually be
found by inspection; but when the term is large, or contains
many factors, it may be advantageous to resolve
it into successive sets of factors, beginning with 270 = 2 x 135
small prime divisors as first factors, and then 3 x 90
taking their multiples, as is shown at the right. 6 x 54
6 x45
57. Eesolve into factors a^ + 33 a? + 270. 9 x 30
270 = 16 X 18 ; 16 + 18 = 33. 10 x 27
.. x^ + 33a; + 270 = (x + 16) (x + 18). 15 x 18
Resolve into factors
58. aj«aj240. 63. a^aj420.
59. a^ + 31 a? h 240. 64. aP + x1260.
60. 2^ + 102^299. 65. a^ + 38a? + 240.
61. 2^2142^480. 66. a^ f 7 a;  1320.
62. «*47aj + 540. 67. a^ + a; — 552.
Case VII.
204. Trinomials having^ binomial factors with unlike terms.
We have considered the factoring of trinomials when their
binomial factors have like terms, as (a ± 6)^, and alscT when
their binomial factors have a common term, as (a? ±a)(x ± 6).
In the remaining case the binomial factors have unlike terms,
and the trinomial has the form of aa^ {bx{c.
205. It follows from § 165, that, when a trinomial is the
product of two binomial factors with unlike terms,
I. The first term of the trinomial is the product of the first
terms of its binomial factors.
II. The third term of the trinomial is the product of the
second terms of its binomial factors.
III. If the third term of the trinomial is positive, the second
terms of its binomial factors will both have the sign of its
middle term.
§ 205.] FACTORTNG. 107
rV. If the third term of the trinomial is negative, the
second terms of its binomial factors will have unlike signs.
V. The middle term of the trinomial is the algebraic sum
of the crossproducts of the terms of its binomial factors.
ft
1. Resolve into factors 6 a?^ — 19 a; f 10.
6x^ = Sxx2x; 10=2x(6); 3 a; x ( 2)+ 2 a; ( 5) = 19a;.
.. 6a;2  19a; + 10 =(3a;  2)(2a;  5).
Since the second term of the trinomial is negative, and the
third term positive, the second term of each binomial factor is
negative.
6a^ = 6 xx^ or 6xx xoT Sx X 2a;, and 10 = 2 x 5 or 1 x 10.
It is seen that 1 and 10 are not the factors to be taken, and that
3 X and 2 x are the only factors of 6 a;^ which by crossproducts
with 2 and 5 will give — 19 x, the second term of the trinomial.
2. Resolve into factors 4:0? \4:ajy — 3y^.
4x^ = 2xx2x; Sy^ = Sy x( y) ; 2x x Sy + 2x x(y)=4xy.
.'. 4x^\4xySy^ = (2x + Sy)(2xy). .
Since the algebraic sum of the crossproducts must be
positive, the greater product must be positive ; and hence
3^ is +, and y is — .
3. Resolve into factors 6 oc^ — ocy — 35 t^.
6x^ = Sxx2x;  S^y^ =  ^y x 7 y ; Sx x( by) + 2x x7 y =  xy.
.'. 67^ xy 352/2 =(2a;  5y)(3a; + 7 y).
Resolve into factors
4. 3ar^22a?4 35. 11. 6a?xi/2if.
5. 6a;2lla?35. 12. 5a? [6xy Sf.
6. 12 m^ 31m + 20. 13. 2 rn? + m?n  3 w?n\
7. 3a?10x{3. 14. 3a^ + ab2b\
8. 3a?^5x2. 15. 10f12ayh2a\
9. 5ar^47a; + 84. 16. 4.0? \4:xy 3f.
10. 3o?'5bx'2l^. 17. 15ar^4a^352r*.
108 ALGEBRA. [§ 206.
206. A trinomial of the form aafi bx + c can be readily
factored by Case VI. if it be first changed into an equivalent
trinomial having unity for the coefficient of its first term.
18. Resolve into factors 3 oj^ — 31 a; + 66.
Multiplying by 3, and dividing the result by 3, we have
It is seen that 3 x is treated as x in Case YI.
19. Resolve into factors 12 aj* + 47 a? + 45.
= (i^^±25^P^±i!)=(3« + 5)(4x + 9).
Since 12 =4 x 3, the first factor, (12x + 20), is divided by 4; and
the second factor, (12 a; + 27), by 3.
Resolve into factors
20. 3ar* + 14a; + 15. 26. 6a^aj40.
21. 7ar^ + 20aj + 12. 26. 6a^13a?46.
22. 5ar^17a; + 12. 27. 3a^ + 5aj12.
23. 6a^\5xe. 28. 3 ar'  17 a? + 20.
24. Sa^'x 10. 29. 10 aj^ + 23 a?  21.
GENERAL METHOD OF FACTORING TRINOMIALS.
Note. The study of this method may be omitted until the pupil
reaches quadratic equations (Chapter XV.).
207. Any trinomial of the form x^ + bx + c may be factored
by changing it to an equivalent expression of the difference of
two squares.
Thus, whatever may be the value of 6, a^ + bx is made a
§ 208.] FACTORING. 109
perfect square by the addition of (  j , for a* f 6a; h [  j
==[« + ] (§ 186). Hence, by adding and subtracting ( „ ) ^
fic* H 6a? + c, we have
"■^HD"a)'=(i)"?
It is evident that if 6* — 4 c is a perfect square, both factors
will be rational.
1. Find the factors of a* + 6 a; + 8.
a;2 + 6x + 8 =(x2 + 6x + 9) 9 + 8 =(x + 3)2  1
= (x + 3  l)(x + 3 + l) = (x + 2)(x + 4).
2. Find the factors of x^ h 9 a; + 20.
3. Find the factors of 3 aj* — 5 aj — 2.
3x25x2 = 3(x2fxi) = 3[xaix+({)afff]
= 3[(x  i)3  S]= 3(x  f + J)(x i J) = 3(x + i)(x2).
4. Find the factors of 2 a?* — 3 maj — 2 m*.
2 x^  3 wx  2 w2 = 2^x2 l^x  mA
o/«. 3m , 6w\/ 3m 6m\ «/ . i n^ ox
= 2^x— + — j^x — — j = 2(x + im)(x2m).
208. In like manner any trinomial of the form aoi^ {bx\c
may be factored. For
\ a aj \_ a \2oy 4 a' aj
_ / & Vy34i^Y 6 Vy4ac\
110 ALGEBRA. [§ 209.
This is a general formula for all of the more important cases
of factoring, and the preceding methods of factoring trinomials
may be considered as special cases under it. It may be used with
advantage in the solution of certain quadratic equations (§ 451).
209. Any trinomial may be directly factored by substituting
for a, h, and c, in the formula in § 208, their values in the
given trinomial.
Thus, in 3aj2  lOa + 3, a = 3, 6 =  10, and c = 3. Sub
stituting these values in the formula, we have
, h , V624ac , 10 , VIOO  36 1
, b Vb^  4 ac , 10 8 «
x{ = x\ = a? — 3.
2a 2a 6 6
Hence 3a^10a? + 3 = 3(aj ^)(x  3).
Find by general formula the factors of
6. 3ar^h5aj12. 8. 3a^13ic + 14.
6. aj2 + 7ajfl2. 9. 2x^ — 5mx — Sm\
7. 2a^ + x2S. 10. 3x^5nx2n\
SUM OR DIPFERENCE OF LIKE POWERS.
210. It follows from § 171, and also from Case V., that the
difference of the even powers of two numbers can be resolved
into two or more binomial factors. Thus,
a^  52 = (a + b) (a  6) ;
a^b*= (a' + b^ (a^  b") = (a^ + 6^) (a + b) (ab);
a^b^ = (a« + b^ (a«  b^ (for factors of a^  b^ see § 212) ;
a«  b'=(a* + b') (a^  b') = (a*+b*)(a^{b^(aAb)(ab)]
and so on.
211. It has been shown in § 173 that the sum of the even
powers of two numbers is not divisible by their sum or difference,
and it will be found by trial that the sum of the even powers
§213.] FACTORING. Ill
of two numbers is not divisible by the sum or difference of
any powers of the numbers except when the eocponents of such
even powers contain an odd factor.
Thus, a* + b^j a* f b\ a^ f b% and so on, are not divisible by
the sum or difference of any powers of a and b ; but a* j 6*
= (ay + (by, and a^^ + &'^ = («")'+ (py, and are each divisible
by a^ + b^y and hence can be factored. Thus,
212. It follows, conversely, from §§ 170, 172, that
a^ + b^=(a{ b) (a^ ab +b^;
a'*  6' = (a  b) (a^ + 06  6^ ;
a? + «»* = (a + &) (a*  a^?^ + a%2 _ a5« + 5^) .
a^ _ 6* = (a  6) (a* + a% + a^^ h a6« + 6*);
and generally that the sum of any two like odd powers of a
and b may be factored by dividing such sum by a + 6 ; and the
difference of any two like odd powers of a and 6, by dividing
such difference by a — b.
213. In like manner many binomials whose terms have
unequal exponents may be factored provided these exponents
have a common factor.
Thus, aj^°— 2/"=(a^*— (2/^*, hence has the factor a^ — T^; and
m* — n^^ = (my — (ny or (m^y — (ny, and hence has as a
factor either m* — n^, or m* + n^, or m^° — n^, or m^^ + n^,
1. Resolve into factors (1) o^ + t/^; (2) a^ — ^.
(1) a;8 4 2/3 =(« + 2/)(«2  ajy + y^).
(2) ix^yi=(^x y)(«2 + ay + yS).
2. Resolve into factors afi — 1/^.
(xfi — y^ = (x^ { y^) (x8 — y8).
a^ + y' = (x + y)(x2  «y + y^) ;
aj8 _ ys _ (a; _ y)(a;2 4. xy + y2).
.. x^  y« = (x + y)(a;  y) (x^^xy + y^)(x^ + ay + y®).
112 ALGEBRA. [§ 214.
S. Resolve into factors 8 a«&»  125 A
v^So^P = 2 a6 ; \/l26c« = 5ca.
. •. (dividing 8 o'd*  126 c« by 2 aft  6 c?)
8a«6«  126c« =(2a6  5c2)(4a«6a + 10a6c2 + 26c*).
4. Resolve into factors (a — 6)' — 8 c*.
^ v^(a6 )« = a 5; v^8^ = 2c.
.'. (dividing a — b  8 c« by a — 6— 2c)
(a _ 6)8 _ 8c« = (a  6  2c)[(a  5)a + (a  6)2c + 4c2J.
Resolve into factors
5. of* + 8. 14. m^n\
6. a^ — 1. 15. m*4w'.
7. l+iB«. 16. ar^1.
8. aj827. 17. l\(a + by.
9. 125iB«. 18. (a 6)8 + 1.
10. 16 — a?*. 19. (m + w)* — (m — n)*.
11. iC*256. 20. (mf w)«(m — n)'.
12. a®27. 21. (2aha;)*(a + 2a;)\
13. a^2/«. 22. Sa^(a\2xy.
214. The sum of any even powers of two numbers, as a^+V,
a* + &*, a® + 6^ (a^" + b^), may be factored by adding to the
binomial such a number as will make it a perfect square, and
then subtracting the number added from the result, as shown
in Case V., §§ 194, 195.
Thus, a* + &* = a* + 6* + 2 a'b^  2 aV
= (a2 {by 2 aV = (a" + b^ + V2a^^ (a* + ft^. VSo^p).
Resolve into factors
23. a;2 + 2/^. 27. a^ + b\
24. a^ + ft*. 28. 9a;* + 42^.
25. m*4w®. 29. 4ar^ + y*.
26. 81 + aJ*. 30. a^4256.
§ 216.] FACTORING. 113
FACTORING BY SYNTHETIC DIVISION.
215. It will be found by multiplying together several poly
nomial factors, arranged according to a common letter or
letters, that the first term of the product is the product of the
first terms of the factorSy and that the last term of the product
is the product of the last terms of the factors.
Thus, (aj2) (a; + 3) (aj5) = aj84aj2 llajf 30; and
(a? 4 a) (x — b) (x + c) =a^ + (a — b + c)a^ — (ab — ac  be) x
— abc,
216. It follows that a polynomial which contains one or
more binomial factors may be factored by resolving its first
and last terms into factors, and then determining by synthetic
division (§ 132) which of the binomials that can be formed by
uniting these factors (two and two) with the sign + or —
are factors of the given polynomial.
This process is made plain below by the factoring of a
few polynomials, and a little practice will enable pupils to
use it readily.
1. Resolve into factors aj" — 4 aj* — 11 a? 4 30.
The three factors of a^ are x, x, and a?, and the three positive
factors of 30 are 2, 3, and 5, or 1, 5, and 6, or 1, 2, and 16.
Since two of the terms of the polynomial are negative, and
the last term positive, try successively by synthetic division the
divisors a; — 2, a? + 3, and x — 5.
1
+ 2
1_4_114.30
+ 2
 4
30
1
3
1216
3
+ 16
16
(x2 _ 2a5 — 15, first quotient)
(x — 5, second quotient)
Hence the factors sought are x — 2, x + 3, and x — 6.
white's alo. — 8l
114
ALGEBRA.
[§ 216.
If we had tried as divisors ajflora? — 1 or a;+2ora? — 3
or x + 5y there would have been in each case a remainder, thus
showing that the trial divisor is not a factor.
In the foregoing process it was assumed that the polynomial x* — 4x>
— 11 X + 30 is composed of three binoinial factors ; but it may be found
on trial that a polynomial of the third degree contains only one rational
binomial factor, the other factor being a trinomial, as in Example 5.
2. Eesolve into factors a:? — 43 a; ■+ 42.
Try X — 1, a — 6, and x + 7.
+ 1
1
+6
l_043+42
+ 1
+ 1
42
1 + 142 (x2+x42,lstquo.)
+6
+42
The process at the left may be
condensed by writing the several
partial products in a horizontal line,
thus:
l_043+42
+1+ 142
+ 1
1
+6
1+142 (x2+x42,lstquo.>
+6+42
1 + 7 (x + 7, 2d quo.)
1 + 7 (x+7, 2dquo.)
The factors sought are x — 1, x — 6, and x + 7.
3. Resolve into factors a?*  10 ic* + 32 ar^  38 a? + 15.
First try in succession x — l,x — 3, x — 5.
110 + 3238 + 15
+ 1  9 + 2315
(x8  9 x2 + 23x  15; 1st quo.)
1
+ 1
1
+ 3
1
+ 6
1_ 9 + 2315
_l 318 + 16
1 6 +
+ 6
5
5
(x26x + 5, 2dquo.)
1—1 (x — 1, 3d quo.)
The factors sought are x — 1, x — 3, x — 5, and x — 1.
4. Resolve into factors 2 ar* + 31 a^ + 62 a?  39.
Try in succession 2 x  1, x + 3, x + 13.
2
+ 1
1
3
2 + 31 + 6239
+ 1 + 16 + 39
1 + 16 + 39
 339
1 + 13
(x2 + 16x + 39)
(X + 13)
The factors are 2 x — 1, x + 3, and x + 18.
216.]
FACTORING.
116
5 . Resolve into factors 3 ic* — 46 05* + 9 a? + 2.
First try 3 xl.
3
+ 1
346+ 9 + 2
+ 1162
115 2
(a;215x2)
If we try aj — 1 or a? + 1 or a? — 2 or a; f 2, we shall have in
each case a remainder, and so conclude that aj* — 15 a? — 2 can
not be resolved into rational factors. This is also made obvious
by the fact that the coefficient of the second term (— 15) is not
the algebraic sum of any two rational factors of — 2 (§ 371).
Hence the factors sought are 3 a? — 1 and a?* — 15 aj — 2.
6. Resolve into factors aj* 3 a?* 18 aj8+ 87 ^— 109 x + 30.
Try first successively x — 2, x — 3, and x + 5.
1
+ 2
1
+ 3
1
6
1318 + 87109 + 30
+ 2 240+ 9430
l_l_20 + 47 16
+ 3+ 642+ 15
1 + 214+ 6
6+16 6
I 13+1
(x2_3«+l)
It will be found by trial that neither a; — 1 nor » + 1 is a
factor of a^ — 3 a? h 1 ; and it is also evident on inspection that
the trinomial has no rational factors.
Resolve into factors
7. aj'f4ar*f aj6. .
8. aj87aj6.
9. aj»19a?30.
10. aj843aj + 42.
11. a^'lOaj^f 31aj30.
12. a^ + 9a^fll«21.
13. aj8a28aj + 12.
14. 3a:»10aj*aj + 12.
15. 2aj89a^ + 13aj6.
16. 6a^+7aj+13.
17. 2a^5aj + 39.
18. aJ*4a^7aj« + 34aj24.
19. aj* + a^4a?16.
20. aJ*~3aj»3ar^ + 12aj4.
I
118 ALGEBRA. [§ 217.
67. a^ — 64. 83. a^ — 27 x \ ISO,
68. 1 — m«. 84. ic' — aj2 — 20a?.
69. 27»'. 85. 4a*14a + 6.
70. l\(x + yy. 86. 4a* — 10a* 44.
71. (xyy\l. 87. 3aj*2a;65.
72. aJ*  12 aj2 + 11. 88. 3a;26a;15.
73. a?* — 8 a;* + 7. 89. a?* — 6 a? — 7.
74. a*10a2 + 9. 90. a:*10aj + 25.
75. a;*42a;120. 91. »«5aj14.
76. a:2 + 21aj72. 92. a?*7a;260.
77. aj2 + a?240. 93. a;* 6ar^ 40.
78. a:* _ 10 aj _ 144. 94. 81 a?*  72 »y + 16 j^.
79. a;2 + 30 aj + 221. 95. a^ + m^a^  n^a^  nM.
80. 21ar^4a;l. 96. a;2^_2aaj + 2aj+(a4 1)'.
81. 36a;231aj56. 97. a*  a^y + af  y^.
82. ar^  41 a; + 420. 98. aJ* (3m  2)a;*  6m.
99. ar^ + 2/* + 2*f2ajy2a»2y2;.
100. a" { b^  (^  <P  2 ab 2 cd.
101. m^ + n^  a^  b^  2mn + 2al).
102. 4 — 4a + 2mii — m^la* nl
103. 9a^{f4:Z^6xySsz4:S^.
104. a^42/^ + l2a;y + 2aj22^.
105. aa^ — by^ + 2 aa^y — 6aj* f aosy^ — 2 bxy,
106. aV2a6a^&^a^aY + 2a62/*&V.
107. ma^ + 2 na^ — nyf — 2 mxy — no? + wiy*.
108. aj39a^ + 23a;15.
109. o^2a^^iea^\2x + 15.
110. a^ + 7aj' + 9a:2 .7aj_10.
§221.] COMMON FACTORS. 119
CHAPTER VIII.
COMMON FACTORS AND MULTIPLES.
COMMON FACTORS.
218. Any number is divisible by any one of its factors or
by the product of any two or more of its prime factors. Thus,
since 30 = 2 x 3 x 5, 30 is divisible by 2 or 3 or 5, or by 2 x 3
or 2 X 5 or 3 X 5. In like manner 3 ab is divisible by 3 or a
or 6 or by 3 a or 3 6 or ab,
219. Two or more numbers are divisible by any common
factor or by the product of any two or more common prime
factors. Thus, since 12^= 2 x 2 x 3, and 30 = 2 x 3 x 5, 12
and 30 are divisible by 2 or 3 or by 2 x 3. In like manner
3 a^b and 6 abc? are divisible by 3 or a or 6, or by 3 a or 36
or ab or 3 ab.
220. It follows that a common factor of two or more num
bers is their common divisor.
Two or more numbers which have no common integral fac
tor except 1 are prime with respect to each other.
Highest Common Factor.
221. The highest common factor of two or more numbers is
the factor of highest degree that will divide each number with
out a remainder. Thus, ± 3 a^b is the highest common factor
of 6 a^bx and 9 a^b^y, and ± 4 a^b^ is the highest common factor
of 12 a^ft^ and  16 a^b\
In arithmetic the greatest number that will exactly divide two or more
numbers is called their greatest common divisor ; but since, in algebra, a
120 ALGEBRA. [§ 222.
negative number is less than any positive number or than a negative
number that is numerically smaller (§ 67), the highest common factor
of two or more numbers may be algebraically less than any other fac
tor, positive or negative. Thus, — 3 a*6 < ± a or ± ah. This explains
and justifies the use of the expression ** greatest common factor '' in arith
metic, and ^* highest common factor '^ in algebra.
222. The highest common factor of two or more algebraic
numbers is their highest common divisor.
For convenience the abbreviation H. C. F. is used in this
chapter for " highest common factor," and H. C. D. for " high
est common divisor."
H. C. F. FOUND BY Factoring.
223. The H. C. F. or H. C. D. of two or more algebraic
numbers may be found by resolving each into its prime factorSy
and then taking the product of all those factors which are common.
The numbers in the following exercises may be readily
resolved into factors by inspection,
1. Find the H. C. F. of 63 a^a?y and ^abn^.
63a2a;83/ = 3 x3x la^o^\
42 ahx^z^ = 2x3x7 abx^z^,
. % 3x7 ax^, or 21 ax% is the H. C. F. required.
The H. C. F. of two or more numbers may be either positive or neg
ative, and hence ± 21 ax^ is the H. C.F. of eSa^hfiy and 42abx^z^, In
this chapter only the positive H. C. F. is given or required.
2. Find the H. C. F. of ar^  0^2^ and aS  2 iB«3/ h ojy*.
ofixy^ = x (pfi y^)z=ix(X'^ y)(x  y);
aj8 _ 2 x^ ^xy^ = x(x y) (x  y),
.'. x{x — y), or x^ _ xy, is the H. C. F. required.
3. Find the H. C. F. of a^12x\S5, aj^10a? + 26, and
aj* + 6 aj — 55.
x2 _ I2x + 35 =(x  5)(x  7) ;
x2 _ lOx + 25 =(x  5)(x  6) ;
x^^ 6x56=(x5)(x+ll).
.*. X — 5 is the H. C. F. required.
§223.] COMMON FACTORS. 121
4. Find the H. C. F. of a*  2 a^h^ +■ 6* and a*  h\
a*  2a262 + 6* = (a2  h^y = ia^  62)(a2  fts) .
.. a2  62 is the H. C. F. required.
6. Find the H. C. F. of 3a^ + 4aj  15 and 5aj*  56aj + 11.
3x2+ 4x16=(3x5)(x + 3)j
6x2  56x + 11 =(6x  l)(x  11).
There is no common factor, and hence no H. C. F.
Find the H. C. F. of
6. 210 and 330; 450, 625, and 825. •
7. 21a86^ and 63a6»c; l^d'ly'c and 60a6V.
8. 2^0? f 7?, IGajy, and 12w^t^.
9. ^a^a?y^, 24:a^ahf^y 72 a»a^/, and 36 aV^^.
10. aj" — ^, a^ — ]^, and a^ — 2xy + i^.
11. aj"  1, (x 1)2, and a^1.
12. a^ 4 aV and aJ* — a^a^.
13. 15 (a?  ly and 45 (a?*  1).
14. 4 (a«  b^ and 20 (a*  b%
15. a^ + &* and a^  a^b f ab\
16. a* — 2aj + l and a^ — 1.
17. a^ + aj + 1 and a^^ + aj^ + l.
18. a^8a^ + 16 and 3a^ 12a^ + 12aj.
19. 10 ay + 40 aV + 40 aV and y*  16«*.
20. aV — ay and ar^ + a^y,
21. aj* 4 8 a? I 16 and a?*  256.
22. a^  10a; + 9, ar*  17a; + 72, and a;^  11 a; + 18.
23. a;* + 10aj21, a^\4:X — 21, and ar* — a; — 56.
24. «»  9a;  36, 2a;*  30a; + 72, and 3a;2  42a; + 72.
25. 3a^4a;7, 5a;»H3a;2, and 15 a;* + 18 a; + 3.
26. a;* + 3;* 132 and a;* + 17 a;* + 60.
122 ALGEBRA. [§ 224.
27. a^llaj»f10a^ and ic* + 7aj — 8.
28. «* — 18aj + 65 and a* + 2aj — 35.
29. 6a? + x — 12 and 4:a^ + 12x + d.
30. a? — ly iC* — 1, and q? — 2qi? + x.
H. C. E. FOUND BY Continued Division.
224. When two or more polynomials are not readily resolved
into factors by inspection, their H. C. F. may be found by the
method of continued division^ a method similar to the corre
sponding one in arithmetic.
225. This method depends on simple principles : to wit,
I. The factor of a number is the factor of any multiple of the
number. Thus, 6, a factor of 12, is a factor of 12 x 2, 12 x 3 •••
12 X n.
II. A common factor of two numbers is a factor of their sum
or of their difference, also of the sum or difference of any of their
multiples. Thus, 6, a common factor of 24 and of 18, is a
factor of 24 + 18 or 2418, also of 24x2 + 18x2 or 24 x
2  18 X 2. Hence
III. A common factor of either of two numbers and their
difference is a common factor of the two numbers. Thus, 6, a
common factor of 18 and of 30 — 18 (or 12), is a common
factor of 18 and 30.
1. Find the H. C. F. of 288 and 816.
288)816(2
676
240)288(1
240
48)240(6
240
48 is the H. C. F. of 288 and 816.
§226.] COMMON FACTORS. 123
Since 48 is a common factor of 48 and 240, it is a factor of
288, their sum (II.).
Since 48 is a factor of 288, it is a factor of 288 x 2, or 576
(I.).
Since 48 is a common factor of 240 and 576, it is a factor
of 816, their sum (II.).
Hence 48 is a common factor oi 288 and 816.
Since any common factor of 288 and 816 is a factor of 240
(II.), and hence a common factor of 288 and 240 and also
of 240 and 48, it follows that 48, the greatest common factor
of 240 and 48, is the H. C. F. of 288 and 816.
226. The following is a general explanation of this method :
Let a and h represent the two numbers, and g, q\ g", and
r, r', r", and so on, denote respectively the successive quotients
and remainders, and suppose that r" = 0. Thus,
h)a{q
r)b(q^
rq^
r')r(q'^
rY
Since r' is a common factor of r' and r, it is a factor of rq'
(§ 225, 1.), and hence of 6, the sum of r' and rq' (II.).
Since r' is a common factor of r and 6, it is a factor of bq
(I.), and hence of a, the sum of bq and r (II.).
Hence r' is a common factor of a and b.
Since any common factor of a and & is a factor of r, and
since any common factor of b and r is a factor of r\ it follows
that the greatest common factor of r and r' (which is r') is the
H. C. F. of a and b.
124
ALGEBRA.
[§ 227.
2. Find the H. C. F. of 6a?7a^^22x+S2 and Saj^+ajlO.
6a;8_7a;2_22x + 32
6«8 + 2aca20a;
3x2 + 3510
2«3
9x22x + 32
9x23x30
3x2 4x10
X42
Sxa + ex
3x6
6X10
6X10
.*. X + 2 is the H. C. F. required.
3. Find the H. C. F. of a^a^b{ab^b* and a^a^hab^^l^.
a*  a^b + ab»  6*
a*  a8&  a262 + ab^
a9  a%  a&2 + 58
a
a^b^  b* = (a^  b^)b^
€fi
 a^b  ab^ + b^
ab^
a^b^
cfi
ab
a^b
+ 68
+ 68
a2  62 is the H. C. F. required.
227. It is seen that a^b^ — b\ the first remainder, will not
divide the first divisor; but aV — b^ = (a^ — b^b^, and, remov
ing the b% we obtain a^ — 6^, which is a divisor. Since b^ is
not a factor of a^ — a^b — ab^ + b% it is evident that it is not
a factor of the H. C. F., and hence may be set aside. At any
stage of the process, a factor of dividend or divisor, not com
mon to both, may be rejected.
4. Find the H. C.F. of 3a^+a^4aj10 and 9a^9x10.
Multiply by 3, 3x8+ x2 4x10
9x8+ 3x2
 12 X  30
lOx
9a;2_9a;_io
9x8 9x2
x + 4
Multiply by 3, 12 x2 
 2X30
36x2
 6X90
(9x29x10)x4 = 36x2
36X40
Divide by 10,
30X50
9a;2_ 9x10
3x6
9x2 16 X
3x + 2
6X10
6X10
.. 3a
;  6 is the H. (
§230.] COMMON FACTORS. 125
6. Find the H. C.F. of 10 aho^ ^ 10 abx^  90 a^x  90 ab
aaid 6 aod^ — 42 aot^ — 36 ax,
10ab3fi + lOabx^  90 a6x  90 a5 = 10ab(iifi ^x^dx 9);
6 aaj*  42 ax^  36 ax = 6ax(sfi  7 x  6);
10 ab = 2 a X 6b ; 6ax = 2ax3x. .*. 2a is common.
x8 + x29x9
x8 7x6
a*_7x 6
x8_2x23x
x87x6
x^ — 2 X — 3, common.
X +2
2x24x6
2x24x6
.. 2a(x2  2x 3) is the H. C.F. required.
228. At any stage of the process, a divisor or dividend may
be multiplied or divided by any factor that is not common to
both, as in Example 4; and any common factor may be set
apart as a factor of the H. C. F., as 2 a in Example 5.
229. To find the H. C. F. of two polynomials by continued
division,
Remove the monomial factor, if any, from ea>ch polynomial^
and set aside the common factor in the same, if any, as a
factor of the H. C, R
Arrange the resulting polynomials in descending powers of
some common letter, and divide the polynomial of the higher
degree by the other, and then the divisor by the remainder (if
any), and the second divisor by the second remainder, and so on
until there is no remainder. The last divisor is the Ei, O. F. of
the first dividend and divisor, and the product of this factor and
the common monomial factor, if any, is the H. C. F. required.
If the two polynomials are of the same degree of the common
letter, either may be used as the divisor.
230. To find the H. C. F. of more ^han two polynomials,
First find the H, C. F. of two of them, and then the JET. C, F.
of this result and a third polynomial, and so on.
126 ALGEBRA. [§ 231.
Find the H. C. F. of
6. a*  18a; + 65 and a^  18aj  35.
7. 6«2 + a;12 and 6a»H7a*aj + 3.
8. a:« + 125 and 2a8 + 7»2 + 75.
9. ar* lOaj 24 and »«_ 23a; 4 28.
10. ar' 16ajt 21 and a;3_j_ 7^ _ 43
11. a:S + 15aj306 and a^26aj60.
12. 4:a^21a^^l5x\20 Biiid a^^6xiS.
13. 2a;«120ajh378 and 5a;«42a;2^73^_,_g3^
14. aj* — a:* + 2a;2 ha;f 3 and a^ + 2ar* — a;2.
15. ar* + 3a* + 4a; + 12 and a;» + 4a;2 + 4aj + 3.
16. 4aJ*8a:320a;2_,_24a? + 20and3a» + 6a*24aj45.
17. a^5ix^\Sa^7x\3 Qnd 2a^9a^^10x3.
18. 18a«51a;* + 13ajf 5 and 6a;2_i3a._5
19. 2a;*3a,'«42a;2_2a;3 and 6ar*«2 + 8aj + 3.
20. a^ — a^y — Qcy^ + y^ and a^ ^ x^y — xy^ — t^.
The H. C. F. of the polynomials in the foregoing examples may be
readily found by synthetic division (§ 216). It is recommended that
the polynomials in several of the foregoing examples be thus resolved
into factors, and their H. C. F. found.
COMMON MULTIPLES.
Lowest Common Multiple.
231. A multiple of a number is the product of the number
multiplied by an integer ; and hence any multiple of a num
ber is exactly divisible by the number. 12 is a multiple of 2,
3, 4, and 6 ; and 3 a& is a multiple of 3, a, and b, also of 3 a,
3 bf and ab.
The product of two or more integral factors is a multiple of each factor.
§ 235.] COMMON MULTIPLES. 127
232. A common multiple of two or more numbers is a mul
tiple of each of them. Thus, 12 is a common multiple of 2, 3, 4,
and 6 ; and 6 a%^ is a common multiple of 6, a', b\ 3 a*, 3 b^,
etc. ; and aj* — ^ is a common multiple oi x + y and x — y,
233. The lowest common multiple of two or more numbers is
the lowest multiple of each of them. Thus, 12 is the lowest
common multiple of 3 and 4, and a^ — ^ is the lowest common
multiple oi x + y and x — y. Hence
234. A common multiple of two or more numbers is exactly
divisible by each of them; and the lowest common multiple of
two or more numbers is the lowest number that is exactly divisible
by ea>ch of them.
For convenience the abbreviation L. C. M. is used for
** lowest common multiple."
The L. C. M. found by Factoring.
235. Since any multiple of a number contains all of its
factors, a common multiple of two or more numbers contains
all the prime factors of each ; and the L. C. M. of two or more
numbers contains oil the prime fojctors of each number, and each
factor in the highest degree in which it occurs.
Thus, the L. C. M. of 3 a^b^ and a%V is 3 x a^ x 6* X c*.
1. Find the L. C. M. of 6 a^ba^ and 10 ab^a^f.
10 a62a;3y2 = 2x6 ab^y^.
.'. 2 X 3 X ^a%Vy\ or ^Oa'^b'h^y^, is the L.C.M. required.
2. Find the L. C. M. of a?f, ^2<x^\f, and 3X^+2 icy+2/l
a;2 _ y2 _ (aj + y) (x  y);
x22a;2/ + y2=(a5y)2;
»2 + 2jcy + y2 = (a; + y)2.
.'. (« + yY X (ac — y)2, or a* — 2 x2y2 ^ y4^ jg the L. C. M. required.
128 ALGEBRA. [§ 236.
8. Find the L.C.M. of a^9, 15 a^  39 a?  18, and Saa^
18 oa? 4 27 a.
x^  9 =(x + S)(x  S);
15x2  39x  18 = 3(x  3)(5x + 2);
3a2x218aa; + 27a=3a(x3)2.
.. 3 a(x  3)a(x + 3) (6 x + 2) is the L. C. M. required.
Find by factoring the L. C. M. of
4. 5a^x and 15 a^ocih/,
5. Uafyn^ and 63an^s^,
6. 6a^c(P, Oa^ftV, 15a6V, and 20aVdaj».
7. 01^ — y^ and a^ — i^.
8. 7x(a+by and 5y^(a^^lf).
9. a?'a% a^ — b% and a* — 5a6 + 4&'.
10. ar*7aj + 10 and a:2__33.__ j^Q
11. a^^ajllO, aj2 + aj90, and 2a? {15x — BO.
12. aj» + 8aj + 15 and aj2 + 4a; — 5.
13. aj2  5aj  84 and a^7aj 60.
14. 7aj2_36a; + 5 and a^25.
15. a^ + od, ab — b^, and a^ — 6*.
16. aj* — y*, a^ — y^, a^ + 7/^, and a?* — 2ajy + 2^.
17. a*  1, a^ + 1, ar^  1, and a^  1.
The L. C. M. found by the H. C. F.
236. When the factors of two polynomials cannot be found
by inspection, they may each be resolved into two factors by
Jiiiding their H, C. F. by continued division, and then dividing
ecich by this H, C. F.
§238.] COMMON MULTIPLES. 129
237. The H. C. F. of two polynomials contains cUl their com
mon factors, and the two quotients obtained by dividing each
polynomial by the H. C. F. contain all the fa/stors not common ;
and hence the product of their H. (7. F, and the two quotients thus
obtained will he the L. C. M, of the two polynomials. It contains
all of their factors, and each in the highest degree.
The process may be somewhat abridged by dividing one of the
polynomials by their H. O. F,, and multiplying the quotient by the
other polynomial,
1. Find the L. C. M. of a^ — a^ ^ xy' + 1^ and a? + a?y
— xy^ — f.
The H. C. F. of the polynomials is x^  j^,
and (pfi — xhf — xy^ + y^^^x^ — tfl^x — y,
,; (x — y) (a^ + x^ — xy^ — y^)m the L. C. M. required.
2. FindtheL.C.M. of or* 48 a? 4 7 and a^ + 8aj* + 9aj + 14.
The H. C. F. of the polynomials is x + 7,
and («848iB + 7)T(x + 7) = x2_7a;4.i.
.. («2  7x 4 l)(x' + 8x2 + oaj + 14) is the L. C. M. required.
Find the H. C. F. and the L. C. M. of
3. 12a^4aj21 and 6a^17aj + 12.
4. aj*23a^10a? and 7a^34ar^4a?5.
5. 3ar»8a^12a;l and 5aj812aj224a;7.
6. So? 5x + 2 Bji^ 21a^20a^llx{ 10.
7. 4aj»3ar*416a:12 and 12aj8 17 a^ + 22 aj 12.
8. ar^2a^a^ and aj» + 2i»2 + 2a; + l.
9. a39aj2f26a;24 and a^6a^ + llaj6.
10. 6aj213a;+6, 2ar* + 5aj12, and 6i»2a;12.
238. The L. C. M. of the polynomials in the above examples
may be readily found by synthetic division, the factors thus
obtained being united as in § 235.
white's jllo. — 9
130 ALGEBRA. [§ 239.
CHAPTER IX.
FRACTIONS.
239. A fraction is one or more of the equal parts of a unit.
The unit which is divided into equal parts is the unit of
the fraction, and one of these equal parts is the fractional unit.
An integer is one or more integral units, and a fraction one or more
fractional units.
S840. A fraction is expressed by two numbers, — one called
the denominator, denoting the number of equal parts into which
the unit is divided ; and the other called the numerator, denot
ing the number of equal parts taken.
241. In arithmetic, fractions are expressed by words or
figures. When expressed by figures in the common form, the
numerator is written above, and the denominator below, a
horizontal line, as ^. In expressing decimal fractions, the
denominator need not be written, but may be indicated by the
decimal point. Thus, seven tenths is written ^ or .7.
In algebra, fractions are expressed by algebraic symbols, the
numerator being written above the denominator, as in arith
metic. Thus, if a unit is supposed to be divided into 6 equal
parts, and a of these parts are taken, the resulting fraction is
expressed by .
In both arithmetic and algebra the numerator and denominator may be
separated by an oblique line (called the solidua)^ as ^/^ and " A (§ 20).
§ 249.] FRACTIONS. 131
242. In algebra a fraction is treated as an indicated division^
the numerator being the dividend^ and the denominator the
divisor. Hence
243. An algebraic fraction is the quotient indicated by writing
the dividend over the divisor, with a line between them.
Thus, the fraction  is the quotient of a divided by b, and
is read " a divided by 6."
244. The numerator and denominator are called the terms
of a fraction. Thus, a and a f 6 are the terms of — ^.
ato
245. Since a5&=acf6c,  = — ; and since acibc=a^b,
^« ^ b be
^ = 2. Hence
be b
Both terms of a fraction may be multiplied or divided by the
same number without altering the value of the fraction. Hence
246. The same factor in both terms of a fraction may be can
cded, or the same factor may be inserted in both terms, without
altering the value of the fraction,
247. Sinceg = ^^(^) =^and:ig= ^^^(^^) =g,
b bx(l) b' 6 _ftx(l) b'
it follows that the signs of both terms of a fraction may be
changed without altering the valu£ of the fraction,
248. 2 or ^^= +% and ±5 or ^^= 2; and hence a
b — b b —b +6
fro/ction is positive when the signs of both its terms are alike, and
negative when the signs of both its terms are unlike. Hence
249. The changing of the sign of either term of a fraction
changes the sign of the fraction.
132 ALGEBRA. [§ 250.
Multiply both terms of
A. H (jL a X OS
1. , — , , , , and by 3.
3 5 m^ m 3
a + b 5 m + 1 a; + y 1+y
Divide both terms of
. 6 9 3a 6 3a „, 3,0
*• IK' To' ~^' To~' ^T' ^^^ ?r~i "y ^'
15 12 6 12a 6 6 9nr
6 ^ 2 > ^ . o , o > and ^ . bya?+«.
a^ — y^a^\'2ocy\y^ x\y ^ ^
a2&2 a3_&3 a* 6* , ,
6. , , and bv a — o,
3a23a6' a'b^ a'2ab + b' y ^ ^'
REDUCTION OF FRACTIONS.
Case I.
250. Fractions reduced to lowest terms.
A fraction is in its lowest terms when its terms contain no
common factor, i.e., are prime to each other (§ 220).
3 a^b^G
1. Reduce — to its lowest terms.
15a6V
3a2&8c Sab^Gxah ab
16a62c8 8a62cx5c2 6(J2
The common factors of the two terms are 3, a, &^, and c, and are
canceled.
a* — 6*
2. Reduce — r to its lowest terms.
a* + 2 d'b^ h 6*
a4 _ 54 ^ (gi 4. 52)(q2 _ 52) ^ q2 _ yi
a*42a262 + 64 (a2  52) (a^ + &2) a^ + ^'
The common factor is a2 4 b^, which is canceled.
S 250.] FRACTIONS. 188
3. Reduce ^1^^ + ^^ to its lowest terms.
a^ — 12 0? 4 35
a;«  lOx + 26 _ (g  b)(x  6) _ a5  6
aj2  12 X + 35 (x  6)(x  7) x  7*
4. Eeduce ^^  ^^^ ^^I'^K to its lowest terms.
a*  a^ft + ab^  b^
(cfia2b^ab^ + J)^)i(a^b^) ^ ab
(a*  a^b + ab*  6*)^(aa  62) a^ab + b^
The H. C. F. of both terms, found by inspection or by continued divis
ion, is a^ — b^; and dividing both terms by a* — 6^ reduce* the fraction
to its lowest terms.
Reduce to lowest terms
6. l^^^. 16. ^ + 2^
25 ab*c a?* + icy + 2^
^ lOS a^b^x ,^ aj81728
144 a& V ^2 _ ^ _ 132
 — 91 m*p^aj* ^ ar*+(a 6)a;4a5
119 mpY^ ' ' aj2+(6hc)a;6c'
g a* — a?h 7? — {a — b)x—db
Sab ' a^ \(b ^c)x — bG
^ 12 afxy ^^ ^t ^
^aa? — 2Aay ' oc^ — y^
10. ^^(^^)' 20. ^'"<
„ aj2 413a; + 22 ^_ a^aaj^
11. — t: _ • m1.
aj2 + 2a?99 a^2aa;+a2
12 .^ + ^^^ . 22. «' + ^
flj8 _ 14iB2 4 49a; a^ + 2a^x + aa?
^3 g» 4 a;  72 ^^ 3 x^ ^ 35 ^jj _^ 105 a;
4aa^48aa;4128a 12 a;^ ^ 132 a;2 _^ 3^0 a;
a«215a? + 54 2 a^ + 19 a; + 35
' a»18ar*4101a;180' * 3 a;^ 4 15 a?  42*
184 ALGEBRA. [§ 251.
a^^ 47 a? 4 14 g^ a? + 125
' a^~54aj35* ' 23^ + 70^ + 76
2a.«13a^ + 23a;12 o^ 16a; + 21
* 7a^33a2^18a; + 8* ' a^ + 7aj48'
a^18a; + 35 a^ + 30;^ + 4fl? + 12
a^55x + 126 ' ix^ ^4.0^ \ 4:X + 3
a*a2a; + 3a^3ar' ^^ a^a^ + 2o^ + x]S
2o. • o«5* — »
4 a^a? — aic^ — 3ic' a^ + 2a:^ — x — 2
29.
a^ 135a; 486 ^^ a^  a^  xf \ f
a;* — 24 a;* — 63 a; — 54 * a;* + a;^ — a«^ — 2/*
261. To reduce a fraction to its lowest terms,
Resolve both terms of the fraction into their prime f actor Sj and
cancel all the common factors; or divide both terms of the frac
tion by their JET. C F,
Case IL
252. Fractions reduced to integral or mixed numbers.
An algebraic mixed number is an integer and a fraction con
nected by the sign + or — . Thus, «^ + j and a^ ^—  are
mixed numbers. ~"
In arithmetic a mixed number is an integer and an added fraction.
Thus, Q\Q^\, and 16f = 16 + 1. In algebra the integral and frac
tional parts of a mixed number may be connected by + or — .
253. Since a fraction is an indicated division (§ 243), it is
reduced to an integer or a mixed number by performing the
operation indicated.
In arithmetic every improper fraction can be reduced to an integer or
to a mixed number, but in algebra this can be done only when at least
one term of the numerator is divisible by a term of the denominator.
Thus, = cannot be reduced ; but
b
= 2 a — + , anA = a + 0.
2a + 6 2a+6 ab
255.] FRACTIONS. 186
1. Reduce — — — ^ — to a mixed number.
= X — a + .
x^ X
2. Reduce ^^ to a mixed number.
JC— 1 05 — 1
Reduce to integral or mixed numbers
^ aa? — bx ^^ a^ — Sx — B
3. . 10. 5 .
X x — l
^ a^ + x ,, a^ — a^ + 3c
4. • 11. .
a a + x
abc — CO? ^o^ + Sajy — 5
5. • 12. : •
ac aj + y
^ a?*3^ „ 3a^15aa^4a'
6. — :; ;;• Xt5. ' ^ •
a^ + .y* 3a? .
a; + 3^ * 4:0^ — aP — 4:X + 1
a^y« a^3a^6 43a6^y
®' aj2^" • a^^2ab + b^
9. — ^ 16.
a — b ic — y
254. To reduce an algebraic fraction to an integral or mixed
number,
Divide the numerator by the denominator.
Case III.
255. Integral or mixed numbers reduced to fractions.
1. Reduce 16^ to a fraction.
i«j 16 X 3 ,2 48 + 2 60
^^ = 3— ^3=3 = T
186 ALG£BBA. [§ 256.
2. Reduce a — r to a fraction
of 6
a« a(a + b)a^ db
d := — ^ ^ •
a + b a \ b a + 6
3. Reduce x^y toa, fraction with x — y for its denominator.
^ . „ _ (a^+y)(xy) _ x^^
x—y x — y
4. Reduce a — h rto a fraction.
gg + ft' ^ (g  b){a + 6)(a« + &') ^ a'  52  ga _ &a _ 26«'
" g + 6"" g + 6 ~ a + 6 a+b
Reduce to a fraction
6. 3i»H. 10. a^ — h^ ^.
6. 6ay + i^^^^. 11. aj«ajy + 3^ ^^
y » + y
7. l_^. 12. {x^zy+ ^
x\'y ^ ' x — z
Sax — a ,«o o Sa^ — 2^
8. 5a ^— 13. Sx — 2y — ^.
x — 1 i» + y
9. a4» V — 14. ; ab + 2.
a\x a{ X
266. To reduce an integer to a fraction,
Multiply the integer by the denominator of the required frao
tion, and under the product wrUe the denominator.
267. To reduce a mixed number to a fraction,
Multiply the integral part by the denominator of the fraction,
and to the prodv/st add, or from the product subtract, according
as the fraction is + or —, the numerator, and under the resvU
ivrite the denominator.
When the fraction is preceded by the sign — , as in Example 4 above,
either its numerator, if a polynomial, must be inclosed in a parenthesis
preceded by the sign — , or the sign of each of its terms must be changed.
§ 259.] FRACTIONS. 187
Case IV.
258. Fractions reduced to their lowest common denominator.
Fractions with unlike denominators may be reduced to equiv
alent fractions with a common denominator, as shown below.
1. Reduce J, , and ^ to equivalent fractions with their
lowest common denominator.
The L. C. M. of 4, 8, and 12 is 24 (§ 234) . Change the fractions to 24ths.
• '• if » if » ^^^ if *^® ^^® equivalent fractions required.
2. Reduce , "!" ,^ , and ^ to equivalent fractions
2a ' 3a* ' 4:a^ ^
with their lowest common denominator.
The L. C. M. of 2 a, Sa^, and 4a8 is 12 a».
3a;6 _ 6a2(3x6) ^ 18q'^36a8 .
2a 12a8 12a8 '
6 + 5a; _ 4a(6 + 6g) ._. 24a + 20fla; .
3a2 12a8 12a8 '
3x 3x3x 9aj
4a8 12a8 12a*
18 a%B — 36 a^ 24 a 4 20 oflj j 9 aj ^v • i 4. *^ *•
. x%?u,^ — !fiiiL ^•'tt r ^v/i*^ gj^^ _^j«/ j^yg ^jjg equivalent fractions
12 a* 12 a8 12 a«
required.
3. Reduce , , and  to equivalent fractions with the
be a
lowest common denominator.
The L. C. M. of b, c, and a is ahc,
a a^c b ab^ c 6c»
6 ~ abc * c ~ abc * a ~ abc
.'. Z, =, and T are the equivalent fractions required.
abc abc abc
269. It is seen from the foregoing solution, that, when the
denominators of the several fractions are prime to each other,
they are reduced to like fractions by multiplying both terms
of each fraction by aU the denomitiators except Us own.
138 ALGEBRA. [§ 260.
4. Eeduce ^"^ , ^~ . and ^ ^ ^ to equivalent fractions
a—ba + b ar^b^
with the lowest common denominator.
The L. C. M. of a  6, a + 6, and a"  &» is a*  6*.
g + & _ (g +6)(a + &) _ (g + ft)^ .
a  6 g2  62 a^b^ '
ab _ (ab)Ca'b) _ (a b)^ ,
a + b g262 a^l^'
3g 3g
aa _ fe2 a2 _ 52
•'• o o » o~ o » ^"^^ T—^^ are the equivalent fractions required.
g2 — 62 a* — 6^ (j2 _ 52
Reduce to equivalent fractions with the lowest common
denominator
 a 6 1 c ^11,1
5. — , — , and —. 10. ,  — , and — — .
be ac ab « — lar— 1 a?l
3c4ac 6a a\bcr — b^a^b
7. IT, :r— r, and • 12.
bx 2x^ lOa? ' a? + 2' a: + 3' aJ' + Sic + e
8. «^, «Jb^, and ;^,. 13.^^^^ ^
3a' 4a2' 6a» * aj3' ajf3' iB«9
9. ?^±^, 5L±6 and ^^±^. 14. ~^^ ^ , ^^
ajy aj*<^ a?y* o^ — y^ 7?\ip oc^ — y^
260. To reduce several fractions to equivalent fractions
with the lowest common denominator.
Find the L. 0. M, of the denominators of the several frdctioiis.
Divide the L. C. M. found by the denominator of each fraction,
and multiply both terms of the fraction by the quotient (§ 245) ;
or multiply both terms of each fraction by those factors of the
L. C. M. which are not contained in its denominator.
§ 262.] FRACTIONS. 139
ADDITION AND SUBTRACTION OF FRACTIONS.
261. In arithmetic only like numbers can he added or sub
tracted; and hence, to add or subtract 5 yards and 2 feet, the
yards must first be reduced to feet, or the feet to the fraction
of a yard; and, to add or subtract f and , both fractions
must be changed to twelfths or to other like fractional unit.
262. The same fundamental principle holds true in algebra;
and hence, to add or subtract algebraic fractions, they must
first be reduced to like fractions, i.e., to equivalent fractions
with a common denominator.
1. What is the sum and the difference of ^ and ^? Of
}a and fa?
The L. C. M. of 8 and 12 is 24. Hence
f  A = if  M = A» difference.
The L. 0. M. of 3 and 5 is 16. Hence
2a, 3a 10a, 9a 19a ^^19^ ^^« ,4a „,^
= = , or — a, or a H , sum.
3 6 15 16 15 ' 16 * 16 '
3a? 4
8a
2a _ 3a ^ 10a _ 9a ^ jflL, difference.
3 6 15 16 16
2. What is (1) the sum and (2) the difference of
andif±^?
12 a
The L. C. M. of 8 a and 12 a is 24 a. Hence
m ^3g4 4a;+3 _ 3(3x4)+2(4a;+3) _ 9a;~1248x+6 _ 17fl;6^
^ ^ 8a 12a 24a 24a 24a
(2") 3g4 4a;3 _ 3C3a;4)2(4x+3) _ 9x128a;~6 _ g18
^ ^ 8a 12a 24a 24a 24a
It must be observed that when a fraction is preceded by the
sign — , as in (2), the entire numerator is to be subtracted, and
hence the sign of each term must be changed or be conceived
to be changed. If the numerator is inclosed in a parenthesis,
as above, the signs of the included terms must be changed
when the parenthesis is removed (§ 104).
140 ALG£BRA. [§ 263.
263. The sum or the difference of algebraic fractions may
be found by connecting them with the proper sign, and then
reducing the result to its simplest form (§ 115).
Simplify
' 5x X ax 0/3^ ^(j?Q^
. 7 ,2 a ^ X 2yx 2ficy'\o^
2x X or y ao aby
g 0^ + 5 a? + 3 ^ 3?y^f 3a;^43y» Gg'a^
xy y * 5 a* 5icV 10 2^
6. 2+. 10. — ^ — ^H ^
y a; X y ^ xy
264. To add or subtract algebraic fractions,
Reduce the given fractions to equivalent fractions with the
lowest common denominator.
Write the sum or the difference of the numerators of these
fractions, as the case may be, over their lowest common denomr
inator, and reduce the result to its simplest form,
, Q. Tij a — 2 , a — 1 a — 3
11. Simplify H •
^ ^ a + 1 a + 3 a'\2
The L. C. M. is (a + !)(« + 3) (a +2).
(a  2)(a + 3)(a + 2) = (a2  4)(a + 3) = «» + 3 a«  4 a  12 ;
(al)(a + l)(a + 2) = (a2_i)(a + 2)=o»42a2_a_2;
(a  3)(a + 3)(a + 1) = (a2  9)(a + !) = («« + a^  9o  9).
a'* + 3 a^ — 4 a — 12, 1st numerator ;
Hence a' + 2 a^ — a — 2, 2d numerator ;
— (a' + a^ _ 9 Of _ 9)^ 3(1 numerator.
a^ + 4 a'^ 4 4 a ~ 5, sum of numerators.
. fl2 gl q3 ^ gS 4. 4^2 _^ 4q _ 5
'a+1 a + 3 a42 (a + l)(a + 3)(a + 2)'
Since the third fraction is preceded by the sign — , the sign of each
term in its new numerator is conceived to be changed when combining
the three numerators.
§264.]
FRACTIONS.
141
Simplify
12. J+ 1
13.
x\y x — y
1 1
a — b a\b
14. ^dbl+^izl.
» — y x\y
15. ^^^_H "^
16.
a^ + V a^b^
1 ^.
x3 aj2
17. ^+ 2
18.
x\4: x — 3
lH2a l2a
l2a l42a'
19 ^ — 1 I ^ — 3 a; + l
a;2 aj2_4 ^^2*
ar — 2 ajf4
20.
21.
aj + 3 a — 5
a {bx a — bx
a — bx a\bx
22. 1+ ^'
a + b a^b^
23.
24.
25.
a
a^
a — 6 a^ — b^
1
1 2a?
aj — a x\a^ x^ — a^
a + b , ab d' + b^
a—b a + b a* — 6*
2B !+«» I liB' 2(l+a^)
a — a; a + a; g? — o^
28. 1^+ ^ ^^
aj + y ^ — y ar^ — 3^
29.
2a
1 3a2
a:^+aajfa^ a?— a o^—a^
30. 1+ /~^„4 "^
31.
a+6 a^^ab+b^ a^+b^
1 a^
»  •
32. . ^ . + ^
33.
34.
a{x — y) x(y — x)
a X
x(a — b) a(a + b)
a 5 a 60
a — 6 o? — b^ a\b
36. ^ + ^^^ *
aj + y 0? — y^ ^\y^
36. _3_+^ + ^^ ^
37.
38.
2a4 82a2 a42
a? 1 , aj + 1
a; — 1 0^ — X
a + b . 2a
+
X
a'
a — b ab + b^
39. 2±b_^_ab_2(^+^^
a — b a + b a^ — b^
.^ m^ ran , n
40. \ •
(m + nf {m+nf m+n
41. ^.— ^+ 2a,
«  3 aj + 3 (aj + 3)^
142 ALGEBRA. [§ 265.
42. J + ^ ^±^ ^^.
x — S x{5 ar* + 2a:15 x + 5
Ao 1 1 6a , gg + Sa
a? — 3a x^3a a^ — 9a^ x — 3a
44. .J_+ 1^,+ «^l
1 , 4aj 5x
45. h
3
46.
x\a {x + ay 0^ — 0?
1 1
aJi^jx+12 a^+aj12'
47. r ;4
aj2_aj20 iB*8aj416 aj2_9a._.20
48. ^^ + ^ 5 + ^^
49.
aja5aj+6 aj28a; + 15 a^7i»+10
1 1
oi^\{a'\b)X'{'ab a^ +{a + c)a; + ac
266. When advantageous, the signs of both terms of a frac
tion may be changed (§ 247), or the signs of an eaen number
of factors in either term, without changing the sign before the
fraction. The sign of an odd number of factors in either term
may be changed if the sign before the fraction be also changed.
Thus, since (6  a) =  (a  6), ^ =r ^ ,
(b — a)(c — a) (a — 6)(a — c)
, a? —X
and
(6 — a)(a — c) (a — b)(a — c)
50. Simphfy ^^ __ ^^^^ _ ^^ + ^^ _ ^^g^ _ ^^ + ^^ __ ^^^^ _ ^y
112
Change to , and — to
^ (6a)(6c) (a6)(6c)' (ca)(c6)
2 ., giving i J + ^
(a — c)(6  c) (a ~ 6)(a — c) (a — &)(6 — c) (^ac)(b—c)
The L. C. M. of the new denominators is (a — 6) (a — c)(6 — c).
. 6c(ac)+2(a6) q& 1
(a  6)Ca  c)(6  c) (a  &)(a  c)(6  c) (a  c)(6  c)
§ 267.] FRACTIONS. 148
Simplify
» a X Ko c^ 5a 6b
61. : r: r' 52.
53.
x(a — b) a(b — a) a — b 6 — a a + b
1 1 1^
(a + by b^  a* (ab)'
s
54. r^, r+ ^
{ah){ac) {ba)(bc)
66. , 4 TT+ ^^
(x — a)(a —b) (a — x)(b — a)
66. ^ + ^ + ^
(xy)(xz) (j/'x){yz) (zx){zy)
a^b , a^b^ ab^l
57. : rrz ZT "i
(a6)(al) (ba)(b + l) (ol)(6 + l)
68. '^y' + y*^ 4 ^
(x+y)(x — z) (y+z)(x + y) (x — z)(y + z)
69. 5±5 + ^±^ + 5±^^ .
a?—(a\b)x{ab a^—{a{c)x\ac Qi^—(b+c)x{bc
MULTIPLICATION AND DIVISION OF FRACTIONS.
266. Since 2xn = — , and ±xn = , it follows that a
b b bn b
fraction may be multiplied by mvMiplying its numerator or by
dividing its denominator,
267. Since ^^n = ^, and 55 ^ n = ^, it follows that a
b b b bn
fraction may be divided by dividing its numerator or by multi
plying its denominator,
1. Multiply ^ by 6; — by m; — by xy.
b my y
2. Multiply — — by ajy; by Aa^; by aW.
if
144 ALGEBRA. [§ 268.
3. Divide^ by 3; ^by3; ^ by ay.
4. Divide ^^^ by 4a^ by Sa^y; by 4aY
6. Multiply 4^8 by a ft; bya + 6; bya«6^
6. Divide 5^^^ by a  6 ; by a + 6; by a* 61
Multiplication op Fractions.
1. Multiply  by .
h n
Since ^ = min, x = xm4n = =^in==^»
n n b b bh
a m _ am
b n bn
3. Multiply 1^ (1) by J^; (2) by ^.
oa — oo a 4 a6
^2N ax + &x gg _ ^ _ x(a + 6)(a2  b^) _ (a + &)g
^'^3a36 a + 6 3(a6)(a + 6) 3
268. It is thus seen tbat the product of two algebraic frac
tions is the product of their numerators divided by the product of
their denominators.
Multiply
4. 1^ by ^. 6. g("'  y) by ^0^ .
5Qcy Say 5a^ 4:(x — y)
2a!y ^ 6oV oft ^ as + y
§ 271.] FRACTIONS. 145
8. "LzJ^ by ^ + ^. 11. ^"^ by ?L±^.
9. ^ by ?^^. 12. ^Il2!by— ^±?
10. $ZJ^ by ^±i^. 13. (^ + yy by ^^.
«^ + 2r « — y (« — y) « + y
269. To multiply a fraction by a fraction,
Multiply the numerators together for the required numerator^
and the denomin^ators for the required denominator, and reduce
the resulting fraction to its lowest terms.
Division of Fractions
270. The reciprocal of a fraction is 1 divided by the fraction
(§ 33). Thus, the reciprocal of f is 1^=, and the reciprocal
of  is 1 ^  = — Hence the reciprocal of a fraction is the
b b a
fraction with its terms inverted.
271. Since —= m!ri, ?s—=^(m*n): and since ? s m
n b n b b
is n times too small, $(m*n) = [hm)xn = — xn = — •
b \b J bm bm
It is thus seen that ^ is divided by — by multiplying ^ by
b n b
the reciprocaJ> of — ; i.e., by — .
n m
1. Divide ^^ by 2"
ab a — b
ah ' ab ab 2 a 2a%'
2. Divide ~ by ^^
xy Q^
3 . 12 _ 3 ^ a;^ _ 3 osV _ ay
xy ' a^ xy 12 12 xy 4*
WHITB*S ALa. — 10
146 ALGEBRA. [§ 272.
3. Divide ?^^^ by ?^=±.
. i = — X — '— = a 4 o.
a+6 a+6 a+6 a—h
Divide
4. 3a^ by i^ 9. (» + y)' by ^ + y
10^^ 5a!j^ » — y (p — y)*
6 6 a'b' (b — c)* 6' — c*
6. ^ by ^('"y) . 11. '^+y' by ^±y
7. ^(fy) by ^i:^. 12. ^"^^^ by £l^.
Sajy xy arHic + l a? + l
8. ^ by ^"^^ 13. ^'^' by ^i^H^.
272. To divide a fraction by a fraction,
Multiply the dividend by the reciprocal of the divisor; or nnuU
tiply the dividend by the divisor with its terms inverted.
EXBRCISBS m MULTIPLICATION AND DIVISION OP FRACTIONS.
Simplify
 ^ ax A: by ^ 3(a + &)^ 46 a —
2b 9a* ^
„ Sa^6a^b^21b(^
« Sx ^ 5a
4.
2 7
3a^ag , 6aa^
T7' 21f'
5^^ 2UdYz
' 3a^y^ Aa^b*(^'
V.
2 '^a + 6 * 6
7.
a — X m? ^ a
a? a'x'' a*a?
8.
X ^ a ^a»<
a{ X a — x Of
9.
a^^b^ ^^a^b" ^^Sa
a — b^ c? 5
10.
6{a^f) _ 16(0! + 2,)
y x — y
§ 272.] FRACTIONS. 147
jj SOCa"  a^ ^ 25(a  a;) ^^ a^y* ^ »*
a* — «* a + 05 aj(a? + y) ' aj(aj* — 2/*)
a^ + ar a + x ar + x — 2 x\2
13. ,?^^?^±^. 18. A+«Yl_2V 6_.
15 <^ + a' . g^y 20 ^^ . ^ + 2^
21. /q^a^ . g'gy42A ^ g'4a?y + y' _
' a^ + 3ic88 a52 + 2aj35 a^{2xS
fa?^9x\lS , a?Sx{15 \ a^Ux\40
* Va5217a?470 'aj«4aj2V a5«3a?18'
a^420a?496 a^8a;20 . a^ + 10g416
' aj*16ajH50 a^ + lOaj24 * oj^Taj + lO'
3^4 17 a? 4 60 3^4 a; 56 . a^H3a?440
'*^ aj2_i4aj 4.49 aj2 + 5aj. 84 * a?7
^ + y^ . /^ ^ — ay + y' y g Y
a?(aa;) a(a4g) 1
^^' a^ + 2ax\a'^a?2ax^a'^cuc
( df^¥ a^^V\ f a^h a^b\
' \a^V a? + v)\ab a\bJ
P* — 2pq 4 g* i>^ 4 2pg 4 g* ' P^ 4 ^^
30. ^ + y' x "^"^^ . a^ + 2rcV4y*
' a^ ^y^ 01^ — a?y^ 4 2^ ' a?* — 2/*
28
29.
148 ALGEBRA. [§ 273.
COMPLEX FRACTIONS.
273. The division of an integer by a fraction, or a fraction
by an integer, or a fraction by a fraction, may be expressed
by writing the dividend over the divisor, with a line between
them.
Thus, 3 s 1 may be written ^ ; I ^ 3,  ; and  ^ f , I
274. A fraction which has a fraction in one or both of its
terms is called a complex fraction.
275. A complex fraction may be reduced to a simple frac
tion by performing the division indicated.
1353^8 24 _, I 272^2 4
 = — ! — = — X — = — ; and ^ — — * — — — ^  — —
15 8 5 5 25' 3^
m. t353^8 24 ,^272^2 4
Thus, •^ = 5 =  x = — ; and 2 = — 5 = x = — •
'458 66 25' 3i 3 2 37 21
276. A complex fraction which contains only a simple frac
tion in one or both terms is best simplified by multiplying both
terms by the L, 0. M, of their denominators,
i i
1. Reduce to a simple fraction (1) ^ ; (2) ^«
2 + ?5
b 2
2. Reduce to a simple fraction (1) j (2) 1 — =•
2a 1+
a a
_A _ & _a(2b+a)
a a
,0.  2 _.. fxq _.. 2a _ q + l2a _ l— o
§ 276.] FRACTIONS. 149
a g — 6
3. Reduce to a simple fraction — =■•
b a — h
g o + & ~ a(a6)+6(q + &) ^ 6Ca2_2,3N aCa^  62)+ 6(a + 6)«
6 06 6(a6) ^ '^
gg  a62 (ggft  2g62 + 6») _ a8  g^ft + ah^  b^
~ g8  ab'^ + g26 + 2 gfta + 68  gS + g^^ ^. aft2 ^ 58
Simplify
a;
6.
1
6.
"g
+ &
1
I* .
1
•*/
a
c
a
a
a&
a
a; 41
14 Jr — !—
X a __ 1 — a
13.
lfa
1J ai 1a
1. a
aJ4 f . 1 14.
^ 1
.
1
1+
a
a?
X —
• —
a
^ a 1
a^b
7. iZl^ t. ^ 16.
oj — a
a? XT i ^~^
x + a
1 1 J I «"!
16. ^ j— 18. —J
616+1 a+1
1. ^^ ^ a?
1a
a + 2f
160
ALGEBRA.
t§ 276.
mSCBLLAIIBOnS BXBRGISRS.
Eeduce to simplest form
1. (^±^.
2.
3.
a»l
(a  1/
g*l'
4. (^^  ^)^
5.
6.
g*l
g«l'
g*a^
(g3ar^(g + a)
7 (g46y(2gf6)«
g + &(2g4&)
8.
9.
ij^ + x 20
a^x12
2aj*H5aj12
4a?9
10. g —
g'
11.
a — b
c^ — a?
12. ^1 ^
«l
aj(a? — 1)
o g , 2g 2g*
lO. r "T"
14.
g — 1 g + 1 g* — 1
X , ___g«
05 — g a^ — a^
15. i2+ 1
16.
17.
g— 1 g g+1
a?h5 ga?42
"y ay
x\2y x — 2y
x — 2y X'\2y
18. A^+ S"' 6
a? + l aj* — 1 a5 — 1
19.
X
f
x — y a? — f a* + ^
20. gj^ %"^X
4 g 5(g + a?)
„ g — aj.ghaj cf — a?
go;
OJ"
4gV
22. ^^±^'x^Il^.
ajy ojy xy
23. g — ^ ^ "^ ^ X ^
a + b a — b a^b
24 a^ + 3^^ 1 x^
a? — 3 aj + 3 aj + 3
25. A+^Ufl^^Y
V g + V V g + V
26. T^F^xfgl^Y
^ / , 2aj \ / 2aj \
V xSj \ xSj
gf 6 a? — y
§ 276.] FRACTIONS. 161
29 a^Sy* 3a?*43y* 6a?^xy' ^
30. « 2y3a? ^ 23/'4a?y + g'
y ab aby
31. y — g a? — ga^ + yg* + afe ^
32. ^ + y^ . a^^ + a^y^a^ + 2r* ^
« — 2^ ' xy
\l + m my\l + m 1 — my
35. (— ^^ + H— ^ i
se /^^ \n m — n\ /m + w wi — n\
\m — n m\nJ ' \m — n m h n/
37. / "c^ + ft a6\ /g + fe I g — & Y
\g — 6 a^bJ \a — b a\bJ
38 / a^ + y a?y \ / a?4y a?y\
\g + 6 "^ g — 6y * \^g — 6 g + 6/
39. f^^^^±lVr^?^^^\
\ X y J \ X y )
40. / "^ I '^ \ ,( '^ I ^\
\^m ■\n m — nj' \m 4 w m — nJ
41. Z'!!?^ n_Vr^^+— ^\
\m — n m\nJ \m + n m — nj
42 ^* — a?* . a^ \Qfi cf — Q^ , g^ — gV + a:!*
g*faJ* g«aJ« g^ + o^ * g* + gV + aJ*'
43. / ^ ^ I ^ V ^ ^
\l4g 1 — aJ'l — a l + g
44 /" a? ^ \^( y _ *^
152 ALGEBRA. [§ 277.
CHAPTER X.
SIMPLE EQUATIONS CONTAINING iPRACTIONS.
277. An equation may be cleared of fractions hy multiplying
both members by the L. C, M. of the denominators (§ 141, II.).
In practice it will often be found convenient to transpose
the terms, and combine those not fractional, before clearing
the equation of fractions, as shown below.
1. Solve the equation 2aj — aj + 5 = aj + 12.
Transposing and combining, 2x — fa; — a; = 7;
multiplying by 12 (L. C. M.), 24x  8x  9x = 84 ;
combining terms, 7 x = 84 ;
whence x = 12.
2. Solve the equation ^±1^Z1^_7 = 6^.
^ 3 5 10
Transposing and combining, ^Jt x — 6 _j_ _« _ ^3 .
clearing of fractions, 10 x + 10  (6 x  18) + 9 x = 390 ;
transposing and combining, 13 x = 362 ;
whence x = 27J.
When the numerator of a fraction preceded by the sign —
contains more than one term, it will be found convenient tc
write the new numerator in a parenthesis, as above, and then
consider the signs changed when combining the terms.
3. Solve the equation — L_ = ^±^.
The L. C. M . of the denominators is x* — 9.
Clearing equation of fractions, 5x + 15— (x — 3) = x + 30;
transposing and combining, 3 x = 12 ;
whence x = 4.
§278.] SIMPLE EQUATIONS. 168
278. When the denominators are partly monomial and partly
polynomial, it may be found advantageous to remove the mono
mial denominator before the polynomial.
4. Solve the equation ^  liSl^l^ = 3  i^.
^ 5 x + 1 10
Multiplying both members by 10, 4 a:  ^C^ ~ ^) = 30  1 + 4 x ;
x+ 1
uraiisi>osing and combining, ^ — \^~ ) = 29
x+1
clearing of fractions,  30x + 60 = 29x + 29;
transposing and combining, — 59x = — 31 ;
multiplying by — 1, 59 x = 31.;
whence ^ = H*
It may sometimes be convenient to combine the fractions in each mem
ber of the equation before clearing of fractions.
Solve tlie equations
5. ^x — ^x = 5. 15. a; » 1(11 — a;) = ^(19 — a?).
6. 20.5 = ^^1. 16. 4 ^
4 8 aj2 aj4
7. x{ix{ix=:ll. 3 2
18 A+?=ii
« «_«o « 2a! a; 12
19. a!26zi£ = ^iai
*• 2 + 3^^ 4
10. a; + ^5 = 4a!17. ^ »
2 3
11. 5_£±i = a!3. ^**' 3ac~6^2»*
,. 2»+3 a;+3 a!4 , „ 21 ^^(a!7)=3.
12. —3 — = ^ + 3. x3
13. 50,^+12 = ^+26. 22. _^ = ^^.
^A ie5 . fi^ 284 a? _„ 12a; + 97 _ 4a; + 16
14. ___+6x = — ^— . 23. ^^p^ ^2
154 ALGEBRA. [§ 278.
24
. ^^gf+^ = 3» + 10. 29. ^+1
x — 2 a^—1 x\l 1 — x
25. 4 = ^ 30. 3  2 _ 1 ^0.
3aj7 3a?5 1a? l + a? 1aj*
a!_6 4_a! ai»9 a; + 3 3x
27. £±l£ri3^1 82. 1 1  ^ ^
a5 — 1 a5 + 3 oj x—2 05—4 a5— 6 a— 8
28 g(«^2) 2(a;3) ^3 33 '^l , »+ l_ 2(a>2)
a! + 2 aj + 3 ' x2'^a! + 2~ iB + 2
84. ^^ + 10 = ^.
35. Sx\(2x + 6)=li{llx32).
36. 2  ^(6aj  4) =i(4a; 18) a.
37 a? +4 •^^ a?23 a?l
4 5 7 *
38. __3a. + 4 =  ^— .
39. i(7a? + 5)(16 + 4aj) + 6 = i(3a? + 9).
40. (3a; + 4)i(16a?)+i(37aj) = 0.
., 17 — 3aj 4a;+2 ^ ^^.7. ,o>,
41. ^ = o — 6x\'(x + 2).
o o o
42.
3a? + 2 2aj4 32a?
43 a?4l a? __ 9 — jc .g 8
03 — 1 aj — 2 7 — a? G — x
^ x'x + l^a^ + x + l^^^
x—1 «+ 1
46. 2a?~ + a^=5a?2a?.
(3a;2)(2a? 3) 6a? 8 2^ . g+lO.
6 5 "^15 3
1 280.]
SIMPLE EQUATIONS.
155
LITERAL EQUATIONS.
279. The known numbers in an equation may be represented
by letters, the first letters of the alphabet being commonly used
for this purpose (§ 11). Thus, in the equation ax]bx = ab,
a and b may represent known numbers ; i.e., numbers regarded
as known.
280. An equation in which some or all of the known numbers
axe represented by letters is called a literal equation.
1. Solve the equation
a
X — b X — a
Clearing of fractions,
transposing terms, .
factoring,
dividing by a — 6,
flwc — a^ = 6a! — 6^ ;
ax — hx = a^ — h^',
(a  h)x = (a — 6)(a + 6);
x = o + 6.
V X
2. Solve the equation  + b = +cL
€L C
Transposing terms,
clearing of fractions,
factoring,
dividing by c — o,
a c
ex — ax== acd — abc ;
(c — a)x = ac(d — 6) ;
ac(d — h)
X = '
c — a
Solve
3. ax+c = bx + 2c.
4. ^ax + ^bx = c.
 CLX ^ mn , J
5. c = [a,
b n
^ a? 05 — 6 d
b. — ^ ^ —•
a
7.
X
X
a+b arb
= 1.
8. — \n = b.
m a
9.  — ^x + c= ^ x — cL
a + b a — b
m — x x^r
10.
x+p
n
X
, a^^ax b^ + bx ^
11. — ; — ! — = a?.
b a
156 ALGEBRA. [§ 280.
ax
X —
a
x^a 7? —
•a'
15.
2+
X
b__
X
d ax + b
X a?
16.
£'
ab
= 1 + 6.
12. 2£zi^ + 65JlA' = a.. 17. oft^o + J
b a xb x + b a^b^
,^ cLX — b aX'\b 5 ,« / , v/ .,v , ..,
^^ ^Ts "^33=^39 '® (p + <^)ip + ^) = {^ + cy.
a 6 6 1^ {x—a)(xb) = {x—a—b)\
14. ^ — = j J.
20. m*aj+2mn— n^=m^+w'.
21. 6w^a?— 2m^=3mn— 4mna?.
22.  +  + =i)^+jpr + gr.
a? a p q r
Probibms.
1. Divide the number 132 into two parts such that one part
may be ^ of the other.
2. Divide the number a into two parts such that one part
may be — of the other.
3. Divide the number 108 into three parts such that \ of
the first equals \ of the second, and also \ of the third.
4. Divide the number n into three parts such that  of the
11 ^
first,  of the second, and  of the third, will be equal.
b c
Let X = first part.
Then ? x 6 = — = second part,
a a
and 2 X c = — = third part.
a a
Hence aj + — +  = n.
a a
Solving the equation, x = — — , first part.
a\b \ c
Whence — = — ^ — , second part ;
a a \b \ c
— = — — — , third part.
a ahbhc
§280.] SIMPLE EQUATIONS. 157
5. Divide the number m into three parts such that  of the
11 ^
first, — of the second, and — of the third, will be equal.
2a oa
6. A man walked 87 miles in 3 days, and ^ of the distance
walked the first day equaled \ of the distance walked the
second day, and ^ of the distance walked the second day
equaled ^ the distance walked the third day. How far did
he walk each day?
7. A father bequeathed ^ of his estate to his eldest son,
■^ of the remainder to his second son, and the rest to his
youngest son; and the eldest son received $1200 more than
the younger. What was the share of each?
8. Divide $735 among three persons so that the second
will have f as much as the first, and the third ^ as much as
the other two together.
9. A man bought a horse and carriage for $ 275, and ^ of
the cost of the carriage plus $ 33 was equal to J of the cost of
the horse. What was the cost of each ?
10. A man bought a horse, saddle, and bridle for $150.
The cost of the saddle was ^ of the cost of the horse, and the
cost of the bridle was ^ the cost of the saddle. What was the
cost of each ?
11. Ten years ago A's age was f of B's age, and 10 years
hence A's age will be f of B's age. What is the age of each
now?
12. At the time of marriage a wife's age was f of the age of
her husband, and 12 years after marriage her age was ^ of her
husband's age. How old was each at marriage ?
13. A man can do ^ of a piece of work in a day, and a boy
can do \ of it in a day. In how many days can both of them
working together do it ?
168 ALGEBRA. [§ 280.
14. A can do a piece of work in a days, and 6 in & days. In
how many days can both together do it ?
Let X = number of days.
Then ~ = part both can do in one day ;
X
^ = part A can do in one day ;
a
 = part B can do in one day.
b
Hence 1 + 1 = 1
a b X
Clearing of fractions, bx{ ax=:ab;
whence x = , number of days.
a\b
15. A can do a piece of work in 8 days, and B in 12 days.
In how many days can both together do it ?
16. A can do a piece of work in 3 days, B in 5 days, and C
in 6 days. In how many days can they together do it ?
17. A and B can do a piece of work in 8 days, A and C in
10 days, and B and C in 12 days. In how many days can A,
B, and C together do it ? In how many days can each alone
doit?
18. A and B working together can build a wall in 8 days,
and A alone can build it in 12 days. In how many days can B
alone build it ?
19. A and B can build a fence in 8 days, and with C's help
they can build it in 6 days. How long will it take C alone to
build the fence ?
20. A man spent J of his money, and then earned ^ as much
as he had spent, and then had $ 21 less than he had at first.
How much money did he have at first ?
21. An estate was so divided between two heirs that J of
the share of the first equaled f of the share of the second ; and
the difference of their shares was $ 362. What was the share
of each ?
§ 280.] SIMPLE EQUATIONS. 169
22. A man paid $ 8100 for two farms, and f of the cost of
the larger farm was equal to ^ of the cost of the smaller.
What was the cost of each ?
23. A piece of carpeting containing 135 yards was cut into
three carpets such that ^ of the number of yards in the first
carpet equaled i of the number of yards in the second, and f of
the number of yards in the third. How many yards were in
each carpet?
24. A cistern can be filled in 4 hours by two pipes running
together, and in 6\ hours by one pipe alone. In how many
hours can the other pipe alone fill it ?
25. Two stoves consumed a certain amount of coal in 12
days, and the smaller stove would consume it in 30 days. In
how many days would the larger stove consume it ?
26. A farmer paid $ 410 for sheep of different grades. For
■J of the whole number he paid $ 10 each ; for ^ of the whole,
$ 7.50 each ; and for the rest, $ 5 each. How many sheep did
he buy ?
27. The sum of $100.50 is contributed by 100 persons.
Some give 50^ each ; ^ as many, 75^ each ; and the rest, $ 1.50
each. How many contributors of each class ?
28. A and B have equal incomes. A lays up ^ of his each
year ; but B spends ^ more than A, and in 3 years finds him
self $ 600 in debt. What is the income of each ?
29. The perimeter of a triangular field is 81 rods, and the
longest side is ^ longer than one of the other sides, and twice
as long as the other. What is the length of each side?
80. A woman sold from her basket \ the number of eggs in
it, and then sold ^ of the eggs remaining, and then had 20
eggs left. How many eggs did she sell ?
160 ALGEBRA. [§ 280.
31. The difference of the squares of two consecutive num
bers is 21. What are the numbers ?
32. A factory employs 200 men, 150 women, and 80 chil
dren, and they receive each week as wages $4110, 3 men
receiving as much as 4 women or 8 children. How much did
each man, woman, and child receive per week ?
33. Divide $ 6000 among A, B, and C, giving A f 300 more
than B, and C one half as much as A and B together.
34. Divide m dollars among A, B, and C, giving A n dollars
more than B, and n dollars less than C.
35. A merchant bought a lot of velvet at $ 1.50 a yard, and
then sold one half of it at $ 2 a yard, one third of it at
$ 1.75 a yard, and the remainder at $ 1.25 a yard, and made a
profit of $ 157.50. How many yards did he buy ?
38. Divide m dollars between A and B in the ratio of a
to b.
Let ax = A^s share,
and hx = B's share.
Then ax + bx = m;
whence x= ^
ahb
Hence ox = «^, A's share;
Ix = 5^, B's share.
a + b
37. Divide $ 120 between two men in the ratio of 3 to 5.
Suggestion. This problem may be solved by substituting f$ 120 for
m, 3 for a, and 6 for 6, in the foregoing formulas, for A*s and B*s shares.
38. The difference between two numbers is 25, and the
greater is to the less as 3 to 2. What are the numbers ?
39. The difference between two numbers is m, and the
greater is to the less as a to 6. What are the numbers ?
§280.] SIMPLE EQUATIONS. 161
40. A boy bought apples at 2 for a cent, and as many more
at 3 for a cent, and then sold them at the rate of 6 for 3 cents,
and gained 11 cents. How many apples did he buy ?
41. A fruit vender bought a certain number of pears at
2 cents each, ^ as many lemons at 3 cents each, and ^ as
many oranges at 4 cents each, and paid $1.96 for the lot.
How many of each did he buy ?
42. A party of 20 persons pay $ 40 for their railroad tickets.
The full fare is $ 2.50, but the children are charged only half
fare. How many children in the party ?
43. A miller made a mixture of barley, com, and oats, using
3 bushels of barley to 4 of corn and 5 of oats. How many
bushels of each grain did he use in a mixture of 72 bushels ?
44. When a colonel tried to draw up his regiment in a solid
square with a certain number of men in the front rank, he had
35 men too many, and when he put one man more in the front
rank he had 30 men too few. How many men in the regi
ment ?
Suggestion. Let x = the number of men in the front rank of the first
solid square.
45. A woman bought a dollar's worth of postage stamps,
receiving a certain number of fivecent stamps, twice as many
twocent stamps less 3, and 3 times as many onecent stamps
less 2. How many stamps of each kind did she buy ?
46. A speculator bought a piece of land at $ 450 an acre,
reserved 5 acres for himself, and laid out the rest in building
lots which he sold at $ 1000 an acre, gaining in the transaction
$11,500 besides the 5 acres reserved. How many acres of
land did he buy ?
47. A man engaged to work for 30 days on the conditions
that he was to receive $1.50 for each day he worked, and
forfeit 50^ for each idle day. At the end of the 30 days he
received $ 27. How many days had he worked ?
WHITENS ALO. — 11
162 ALGEBRA. [§ 280.
48. A man engaged to work for m days on the conditions
that he was to receive a dollars for each day he worked, and
forfeit b dollars for each idle day. At the end of m days he
received n dollars. How many days had he worked ?
49. A farmer can mow a field in 12 hours, his oldest son in
16 hours, and his second son in 18 hours. In how many hours
can the three together mow it ?
50. A, B, and C can do a piece of work in 20 days ; A and
B can do it in 40 days; and A and C in 30 days. In how
many days can each alone do it ?
51. A man being asked his age replied that f of his age 10
years ago is equal to f of his age 10 years hence. What was
his age ?
52. At what time between 2 and 3 o'clock are the hands of
a watch together ? At what time between 4 and 5 ?
53. At what time between 3 and 4 o'clock are the hands
of a watch opposite each other? At what time between 8
and 9 o'clock ?
54. A man spends one fifth of his yearly income for house
rent, one half of the remainder for provisions, two fifths of the
remainder for other expenses, and lays up $ 240. What is his
yearly income ?
55. A steamer, running 18 miles an hour, follows a ship 16
miles off, that is sailing 10 miles an hour. How many miles
must the steamer run to overtake the ship ?
56. A courier starts from a certain place and travels at the
rate of 10 1 miles an hour. Two hours later a second courier,
traveling 13 1 miles an hour, is sent to recall the first. In how
many hours will the second courier overtake the first ?
57. A person has just 5 hours at his disposal. How far
can he ride with a friend in a buggy, going 10 miles an hour,
and walk back at the rate of 4 miles an hour ?
§282.] SIMPLE EQUATIONS. 163
GENERAL PROBLEMS.
281. When the given numbers in a problem are represented
by letters, it is called a general problem ; and its solution is a
general solution for all problems of that class. It also serves
as a model solution. Several of the foregoing problems are
general problems.
282. The result obtained by the solution of a general prob
lem is called a solution formula, or, briefly, a formula ; and, by
substituting for the letters in such formula the particular num
bers given in a similar problem, the numerical answer to such
problem is obtained.
1. The time past noon is n times the time to midnight.
What is the time of day?
Let X = time past noon ;
then 12 — X = time to midnight.
Hence « = n (12 — x).
Transposing terms, x\nx=l2n;
12 n
whence x = — —, time past noon ; (1)
1 f w
12  X = 12  i?^ = ^^, time to midnight. (2)
1 + w 1 + n
2. The time past noon is f of the time to midnight. What
is the time of day ?
Let x = time past noon, and substitute  for n in Formulas (1) and
(2) above, and we have
X = i^ = ^iii = 3d = 44 = 4h. 48 m., tune past noon.
1 + n I i
12  aj = i? = — = 7i = 7 h. 12 m., time to midnight.
3. A man, being asked the time of day, said that f of the
time past noon equals  of the time to midnight. What was
the hour of day ?
164 ALGEBRA. [§ 282.
4. What is the time of day when ^ of the time to noon is
equal to ■§• of the time past midnight ?
6. A courier pursues a second courier, who has a start of m
miles ; and the first courier travels at the rate of a miles an
hour, and the second at the rate of b miles an hour. In how
many hours will the first courier overtake the second ?
Let X = the number of hours ;
then ax = distance traveled by first courier,
and bx = distance traveled by second courier^
Hence ax—bx = m;
whence x = ^* , number of hours.
a — b
6. An express train running 36 miles an hour follows an
accommodation train running 24 miles an hour, and the accom
modation train has a start of 84 miles. In how many hours
will the express train overtake the accommodation train ?
Let X = number of hours.
Since « = — ^L_, and m = 84, a = 36, and 5 = 24,
a — b
84
X = = 7, number of hours.
3624
7. A train sets out from A for B, which is n miles distant,
running a miles an hour ; c hours later another train leaves B
for A, running b miles an hour. How far will each train have
run when it meets the other ?
Let X = the number of hours run by first train ;
then ax = distance run by first train ;
X — c = number of hours run by second train ;
b(x — c) = distance run by second train.
Hence ax + b(x — c)=n;
whence x = ^±^ ;
a + b
ax = "(^ "^ ^^) , distance run by first train ; (1)
o + 6
6(x — c) = ^t^.Il^£2, distance run by second train. (2)
§282.] SIMPLE EQUATIONS. 166
8. The railroad distance from Buffalo to Chicago is 540
miles. An express train leaves Buffalo for Chicago, running
at the rate of 35 miles an hour ; and 5^ hours later an express
train leaves Chicago for Buffalo, running at the rate of 40 miles
an hour. How far will each train have run when they meet ?
SuGGESTioy. Solve this problem by the use of the general formulas
(1) and (2) obtained in solving Problem 7.
9. Divide the number m into four parts such that the first
increased by w, the second diminished by n, the third multi
plied by w, and the fourth divided by w, will be equal.
Let X = number to which the results are equal ;
then x — n — first part ;
jc + n = second part ;
 = third part ;
n
nx = fourth part. ,
Hence a!n + « + n + ?+ na! = «.
n
Combining terms, 2a544 wx = wi;
n
clearing of fractions, 2nx\x\ n^x = mn ;
factoring, (»^ + 2 w + l)a! = mn ;
whence x = ^^
(n + 1)2
35 — n = — ^^ ^ — n, first part ;
(n + 1)2
x+n = — ^^^ 1 n, second part ;
(n + 1)2
* ^ . third part ;
n (n + 1)2
nx = , ^^ — , fourth part.
(n + l)2
10. Divide the number 80 into four parts such that the first
increased by 3, the second diminished by 3, the third multi
plied by 3, and the fourth divided by 3, will all be equal.
SuoGEBTiON. Let x denote the number to which the several results
are equal, x^S denoting the first part, a; + 3 the second part, and so on ;
or solve by substituting in the foregoing formulas.
166 ALGEBRA. [§ 283.
11. Divide the number 100 into three parts such that the
first increased by 8, the second diminished by 8, and the third
divided by 3, will be equal.
12. Divide the number m into three parts such that the
first increased by n, the second diminished by n, and the third
divided by n, will be equal.
13. The sum of two numbers is m, and their difference is
n. What are the numbers ?
Let X = the larger number ;
then X — n = the smaller number.
Hence x\x — n = m.
Transposing terms, 2 as = w + n ;
x=™±»,thelargernnmben 0)
X  n = ^^^±^  n = ^Lm^, smaller number. (2)
2 2 ^ ^
283. It is seen from these two formulas that when the sum
and the difference of two numbers are given,
I. To find the larger number, add the sum and the difference
of the two numbers^ and divide the resuU by 2,
II. To find the smaller number, subtract the difference of
the two numbers from their sum, and divide the result by 2,
284. In like manner the formula obtained by the solution
of any general problem may be expanded into a rule for the
solution of all like problems.
Such a formula may also be used for the solution of problems that vaiy
somewhat from the general problem from which it is derived.
14. Expand into a rule the formula reached by the solution
of Problem 1.
15. Expand into a rule the formula reached in the solution
of Problem 5.
§288.] SIMPLE EQUATIONS. 167
285. Percentage formulas. — The problems in simple percent
age involve three numbers such that, if two are given, the third
may be found. These numbers may be represented by letters,
as follows :
h = base, or the number of which the per cent is taken,
r = rate per cent, or the number of hundredths.
p =z percentage, or the number found by taking the given per
cent of the base.
Since p = hr, r =^j and 6 = — , and hence the formulas :
r
(1) p=br. (2) r = . (3) 6 = . (4) 6=^
By these formulas all the problems in simple percentage
may be solved.
16. If 5% of a certain ore is silver, how much silver is
there in 3740 pounds of the ore?
17. 2400 pounds of iron ore yielded 1008 pounds of iron.
What per cent of the ore was iron ?
18. If copper ore is 25% copper, how many pounds of the
ore will yield 1000 pounds of copper ?
286. Interest formulas. — The problems in interest involve
five numbers or elements : the principal, the rate, the time, the
interest, and the amount. These five numbers are so related,
that, if any three are given, the other two may be found.
287. These numbers may be represented by letters, as fol
lows:
p = principal, or the money on which interest is computed.
r = rate per cent, or the number of hundredths.
t = time, expressed in years or parts of a year.
i = interest, or the money paid for the use of the principal.
a = amount, or the sum of principal and interest.
288. The interest on any sum of money for one year is
found by multiplying the principal by the rate per cent, ex
168 ALGEBRA. [§ 289.
pressed as hundredths; and the interest for any given time
is found by multiplying the interest for one year by the
number of years. Hence i=prt. '
• • •
Since i = prt, r = — , t= — , and p = —; and we thus obtain
pr pr rt
the following special formulas for the solution of the several
classes of problems in interest :
(1) i^prt, (2) r=±. (3) t = ±^. (4) p = l
19. What is the interest of $ 160.80 for 2 yr. 3 mo. at 8 % ?
20. The interest of $ 95.40 for 3 yr. 9 mo. is f 28.62. What
is the rate per cent ?
21. The interest of $56.78 for a certain time at 10 % was
$ 22.24. What was the time ?
22. What principal will produce $86.80 of interest in 2yr.
4 mo. at 6% ?
289. Since the amount is the sum of principal and interest
(a = j9 h t), a=p ^ptv] and hence, factoring, a =i)(l 4 rt),
and p= Again, since a=p\ptr, a—p=:ptr\ and
hence r = , and t = —
pt pr
We thus obtain the following formulas :
(1) a=p + i=p+prt. (3) < = ^
a—p /A\ ^ — ^. *• ^
(2) r = ^^^. (4) p = at =
pt ^ ' l{rt
23. A note of $95.40 Sit S^oj when paid, amounted to
124.02. How long did the note run ?
24. A note bearing interest at 9 % amounted in 3 yr. 2 mo.
to $ 360.75. What was the face of the note ?
Note. For additional problems in percentage and interest, see White's
'«New Complete Arithmetic,*' pp. 178192 and 240246.
5 293.] SIMULTANEOUS EQUATIONS. 169
CHAPTER XL
SIMULTANEOUS EQUATIONS.
SIMPLE EQUATIONS WITH TWO UNKNOWN NUMBERS.
290. In a simple equation with only one unknown number,
there is one value of this number, and only one, that will
satisfy the equation. Such an equation is said to be determinate ;
i.e., the unknown number in it has a definite value. Thus, (1)
3 a; = 12, and (2)  a; = 10, are determinate equations, the value
of X in (1) being 4, and in (2), 15.
291. When an equation contains two unknown numbers, as
X { y = 12, an indefinite number of values of these unknown
numbers will satisfy the equation, and for this reason the
equation is said to be indeterminate. Thus, in a; + y = 12, the
value of X will vary with the value of y, and the value of y will
vary with the value of oj. If ^ = 4, a; = 8 ; it y= 2, a? = 10 ;
if aj = 7, y = 5i if x = W, 2/ = — 3; and so on.
292. Two equations that express different relations between
two unknown numbers are said to be independent. Thus,
x\y = 12, and ocy = 35, express different relations between x
and y, and hence are independent equations.
Independent equations cannot be made to assume the same form.
Thus, x + y = 12 (1), and 2 a; + 2 2/ = 24 (2), can be made to assume the
same form by multiplying each member of (1) by 2, or dividing each
member of (2) by 2, and hence these equations are not independent.
293. The unknown numbers in two independent equations
may have the same value in both ; and, when this is the case,
the substitution of these values for the unknown numbers will
satisfy both equations simultaneously. Thus, the values of x
170 ALGEBRA. [§ 294.
and y in a; — y = 2, and xy = 35, are respectively 7 and 6, and
the substitution of these values satisfies both equations.
294. Two independent equations in which the two unknown
numbers have each the same value are called simultaneous
equations. Thus, a; + 3/ = 13, and a; — ^ = 3, are simultaneous
equations.
Elimination.
295. The solution of two simultaneous equations involves
their combination in such maimer as to remove or eliminate
one of the unknown numbers.
Thus, let 2 a; + 1/ = 42, and 3 a? — 2^ = 33, be two simultaneous
equations. Adding the first members and the second mem
bers of the two equations (Ax. 1), we have
2x + y = 42 (1)
3a?y = 33 (2)
5 a; =75 .. a? = 15.
Substituting 15 for x in (1), we have 30 + ^ = 42. .. y = 12.
296. The process of combining two simultaneous equations
in such manner as to obtain a single equation with only one
unknown number, is called elimination.
297. There are three general methods of elimination : to wit,
I. By addition or subtraction.
II. By substitution,
III. By comparison
298. I. Elimination by addition or subtraction.
1 . Find the values of x and y in the equations
5a; + 4y = 40, (1)
7x2y = lS. (2)
Multiplying both members of (2) by 2, 14x — 4 y = 36 ; (3)
adding (1) and (3) member to member, 19a; = 76 ;
whence x = 4.
{
§298.] SMlXTAXEOrS EQCATrOX?.
in
SnlEtifcmiag^iarziiL :i ^
21)^4^=40;
tnm^msing, etc.
4j=a>;
whence
J = ^
Hence 2=4, and y=ri; and tboe vaioBB of jr and y w€l aalas^ tj^
giren equatiaBS.
2. Solve tbe eqoatioiis
Midtiplyiiig (1) bj 2, and (2) br 3»
8iu6j = ll\
v3^
15xh6jr = ia>;
K*^
sabtiactiiig (3) from (4),
7x = 70;
whence
x = 10L
Sahsthnting 10 lor x in (1\
4O + 3y = o0;
tnuisposing;
3jr = lo;
whence
f = a
M=^
0^
3. Solve the eqnatiann
5 ' 3
vn
Clearing (1) and (2) of fractions,
5x + 4> = U!iX
vSX
9x + 10yS30;
v«^
multiplying (3) by 6, and (4) by 2,
25i + 20y = !»0.
V''^
18i + 20y = tS»iO;
v«»^
subtracting (6) from (5),
7* = l»0;
whence
a! = 20.
Substituting 20 for x in (3),
100 + 4y = U50;
transposing,
4y = t50;
whence
J»=15.
It is sometimes not necessary to clear the equations of fractions, since
the coefficients of one of the letters may otherwise be made the same, as
shown in the following solution.
i
172
ALGEBRA.
[§ 298.
4. Solve the equations
Multiplying (1) by 2,
adding (2) and (3),
whence
Substituting 8 for x in (1),
transposing,
Solve the equations
( x + y = lS,
^' 2ajy = 21.
raj42y = 20,
(4:x3y = 13,
^' (6aj4y = 22.
®' l5x2y = SS.
(xh3y = 36,
Xx2y = 16.
1 7a; 32^ = 62.
(3x + 4.y = U,
^^' \^y_Sx= 1.
9.
10.
12.
3a; + ^ = 28,
5aj^ = 37.
2
6a 2
8 6
= 6,
4 3
££_? = 4.
5x^
4 3
§^ or2x
x =
«+l
12;
16;
8.
6;
(1)
(2)
(3)
V
6
? = 1. /. ysA.
IS, ^
= 9,
«_3/
3 6
^^ = 1.
14. <
( 2x
5
3x
I 8
32
4
3y
15. ^
T + 6
x.y
8 6
= 6,
= 6.
14,
4.
16. ^
3a; n — = Si«
§299.] SIMULTANEOUS EQUATIONS. 173
To eliminate by addition or subtraction,
Multiply or divide the given equations by such numbers as wiU
make the coefficients of one of the unknown numbers equal in
the resulting equations.
If the equal coefficients have unlike signs, add the resulting
equations; and, if they have like signs, subtract one equation from
the other.
Note. The adding of the first members and the second members of
two equations is called briefly the adding of the equations; and in like
manner the subtracting of the first and the second members of one equa
tion respectively from the first and the second members of another equa
tion is called briefly the subtracting of one equation from the other.
299. II. Elimination by substitution.
17. Solve the equations \j^^V^^'^^' 9^
^ \2x3y= 5. .(2)
Eliminate x.
From (1) we find
a; = 202y.
Substituting 20  2 y for » in
(2),
2(20 2y) 32/ = 5;
simplifying,
407y = 5;
transposing,
7y = 35;
dividing by — 7,
y = 5.
Substituting 5 for ^ in (1),
a; + 10 = 20 ;
whence
X = 10.
18. Solve the equations
(6y3aj = l, (1)
l3» + 42/ = 37. (2)
Eliminate y.
From (1) we find
,.5^
Substituting 5*^ for y in
5
(2).
5
clearing of fractions, etc..
27a; = 189;
whence
« = 7.
Substituting 7 for x in (1),
6y21=l;
whence
y = 4.
174
ALGEBRA.
[§300.
Solve the equations
19.
20.
21.
22.
23.
24.
i2x^3y = 22,
j2x + 3y = 27,
(4aj— y= 5.
(2a; + 5^^ = 19,
( x7y= 0.
(5a? + 22/ = 29,
\2y x = l.
( y
(5a;
25.
f4a; + 3y =
\2y4cx =
22,
12.
26.
I
33.
4
= 10,
= a
27. <
22!=6,
2^422 = 34.
5a;32^ = 34,
3a;= 2.
28.
2^3 '
 + ^ = 74
2 "^ 2
2a;32/ = 0.
= 6,
To eliminate by substitution,
Find from one of the given equations the value of the unknoum
number to be eliminated.
Substitute this value for the unTaiown number in the other
equation, and solve the resulting equation,
300. III. Elimination by comparison.
29. Find the values of x and y in
{
Eliminate x.
From (1) we find
and from (2),
solving the equation,
Substituting 6 for y in (1),
whence
a; + 2t/ = 25,
(1)
2x3y = lb.
(2)
x = 252y;
(3)
^ 16+3y
* 2 •
(4)
(3) and (4), 262y=i^^;
y = 6.
X + 10 = 26;
« = 16.
§300. J
SIMULTANEOUS EQUATIONS.
175
30. Solve the equations
Eliminate y.
From (1) we find
and from (2),
Equating the values of y,
solving the equation,
Substituting 18 for z in (1),
whence
4 + 3~^'
(1)
y 2 .,
.3 6~
(2)
V = 36^;
, = 3+1
8+1 = 36^;
a = 18.
1+6 = 9;
tf = 12.
Solve the equations
31.
32.
33.
34.
35
36.
37
38.
i2x y
X + 2Z:
2ySz:
4:y — 5z:
5 X{4:Z.
ilx^z
{
{
{
{
{
{
y 25 =
3^ + 52 =
4:X + 12y
X+ 6y
y2z =
4:ySz =
x + Sz =
3x'2z =
= 16,
= 36.
= 18,
= 8.
= 9.
= 11,
= 15.
= 4,
= 20.
= 5,
= 2.
6,
26.
50,
7.
39.
40.
41.
42.
43.
I
i+^=>*
^ y ^
X—Z = 4:y
Sx , 2z K
T+3 = ^
L~3 4 ''*•
{
ax — by = 10,
ax\by = 26.
4 + T"^^'
^2 T"" ^^
176 ALGEBRA. [§301.
To eliminate by comparison,
Find from each of the given equaiiona the value of one of the
uiiknown numbers in terms of the other.
Equate the values of the unknovm number thus founds ana
solve the resulting equxUion.
MISCBLLAIVBOUS EXBRCISBS.
301. The following pairs of simultaneous equations may be
solved by any one of the above methods of elimination. The
forms of the given equations will usually indicate which
method will be the most advantageous.
«
302. It will sometimes be best to simplify the equations
before determining the method of elimination to be used.
Thus, the equations
(1) ^_?LzJ^ = 2,and(2)^+^ = 10,
may first be changed (1) to x{lly = 60, and (2) to
5 a? + 2/ = 60, and then either x or y may be eliminated by
subtraction or by substitution.
The method by addition or subtraction will usually be found the most
advantageous ; but, when the coeificient of one of the unknown numbers
in either equation is unity, the method by substitution may be preferable.
Solve the equations
^ (x + y = 19, ^ <llx + 3y =
\x^y=: 7, '(4a? — 7y =
2 <2x^Sy = lS, g <15x13y =
\5x + 4ty = 22, ' (17a? y =
= 100,
4.
= 21,
65.
3 ( 7a?+ 42/ = 85, (Sx5y= 5,
(I3aj37y = 69. ' (5a;3y =
3y = 36.
§302.]
7.
8.
9.
10.
SmULTANfiOUS EQUATIONS.
177
11.
12.
31.
Xl9x21y = 
<25x^21y= 410,
Xl6x14:y = 12.
14:X13y= 51,
152^— a; = 123.
xj2y = 3,
5y{4LX = 6.
13.
14.
I
I
J
15.
16.
2 3
5 + 2 = 8.
13 2
17.
18.
f60a;17y = 146,
(48a; + 13y = 170.
f4xf12y = 5,
( aj+ 2y = l.
f ia:f 3y = 8,
(K3a:2)=2y + 3.
2ajy= 7.
7 3
a? — y = 4.
19.
20.
21.
22.
ii(^ + y)K^y)= 2,
U(«^ + y)+i(x2,)=lo.
((» + 5)(2^ + 7)=(« + l)(y9)+112,
I 2a; + 10 = 3v + l.
23
24
25
26
27
(5f2^lla; = 4y + 117,
I 8a; + 175 = 2y.
f 49a?  37f + 4Sy = 9x + 273,
X 3x26{ = 5y.
13a;472^ = 121,
2x + \y=. 14.
10^a;52/ = 40,
9a; 8.52^ = 20.
.2a; + .3y = 4.3,
.2 a? .2 2^ = 1.5.
1.7 a;  2.3 y = 1.6,
.4 a; + .06 y = 2.18.
.2 a; 4 .4 2^ = 2.8,
.4 a; + .3 2^ = 3.1.
whitb's alo. — 12
28. <
29.
1
4 + 4 ^t'
x + 2 y + 2 _..
6 6 ~
a
a;
30.
a;f2^=»,
la? — v = d.
IT
«.
— . = «
» ft
?^ = « —
^r=<»
33.
X
n
*
I
m
37.
2,
X
34.
f^ 2_= 1
)arh a — h ^
1 aj ^ a — h
X
303. When simultaneons equations are fractional, and of the
form ±= Cy they may be solved most readily by treating —
X y
and  as the unknown numbers.
y
\ X t/
89* Solve the equations < ^^ ^
Kliminate z.
Multiplying (1) by 8,
multiplying (2) by 2,
Nubtracting (4) from (8),
Hubitituting 8 for y in (2),
^ + ?l = 18;
» y
?1 = 3.
(1)
(2)
(3)
§304.]
SIMULTANEOUS EQUATIONS,
179
Solve the equations
40.
? + ? = !,
X y
^ 2J y
41. <
42. <
1* 3_3
X y ^
ay
rl , 1__ 8
X y JlO
112
a;
2^ 15'
43. <
cm , ?i
— +  = «,
 + — = 6.
La? y
45.
46.
44. <
X y
5
6'
2_3
5
6
7
a;
y
47. <
^aj y
X y
n m
X y
a? 2/
a? 2/
1
6^
23
21 '
= P,
EQUATIONS WITH THREE OR MORE UNKNOWN NUMBERS.
304. When three equations with three unknown numbers
are given, we may combine one of the equations with each of
the other two in such manner as to obtain two equations with
two unknown numbers. The two resulting equations may then
be combined, as in the preceding cases, giving a single equa
tion with one unknown number. The value of this unknown
number being found, the value of the other unknown numbers
may be obtained by substitution. This process may some
times be shortened, as shown in Example 2.
1. Find the values of x, y, and z in
3aj + 52^+ 2 = 26,
6a; + 3^ + 32 = 36,
9x + 4:y + 4:Z = 50.
(1)
(2)
(3)
180 ALGEBRA. [§ 305.
First eliminate z.
Multiplying (1) by 3, 9 x + 16 y + 3 a = 78 ; (4)
subtracting (2) from (4) , 3 x + 12 y = 42. (5)
Multiplying (1) by 4, 12 x + 20 y + 4 5? = 104 ; (6)
su^t^acting (3) from (6), 3 x + 16 y = 64. (7)
Next eliminate x from (5) and (7).
Subtracting (5) from (7), 4 y = 12. .. y = 3.
Substituting 3 for y in (5), 3 x = 6. .•. x = 2.
Substituting 3 for y, and 2 for x, in (1), « = 6.
Hence x = 2, y = 3, « = 5.
r2aj + 32/+ « = 25, (1)
2. Solve the equations ^ 4 a; + 5 1/ + 2 « = 46, (2)
(60? + 42^ + 42; = 58. (3)
Multiplying (1) by 2, 4 x + 6 y + 2 « = 60 ; (4)
subtracting (2) from (4), y = 4.
Multiplying (1) by 3, 6 x + 9 y + 3 « = 76 ; (6)
subtracting (3) from (5), 6 y — « = 17 ; (6)
substituting 4 for y in (6), z = 3.
Substituting 4 for y, and 3 for «, in (1), 2 x = 10. . •. x = 5.
305. When any one of the unknown numbers does not occur
in all three equations, we may first eliminate such unknown
number from the equations in which it does occur, thus obtain
ing two equations with two unknown numbers.
(x\y =
11,
(1)
3. Solve the equations \x\% =
:10,
(2)
(y + z =
9.
(3)
First eliminate x from (1) and (2)
Subtracting (2) from (1),
y — a = l.
W
Adding (4) and (3),
2y = 10.
.•. y = 6.
Substituting 6 for y in (3),
« = 4.
Substituting 6 for y in (1),
x = 6.
These equations may also be solved by first adding (1), (2), and (3),
then dividing the resulting equation by 2, and from this equation sub
tracting successively (1), C2), and (3).
§306.] SIMULTANEOUS EQUATIONS. 181
306. When four or more equations with four or more un
known numbers are given, one of the unknown numbers may
be eliminated by combining one of the equations in which it
occurs with each other equation in which it occurs ; then a
second unknown nunxber may be eliminated by combining one
of the resulting equations in which it joccurs with each other
resulting equation in which it occurs ; and so on.
When each imknown number does not occur in all of the
given equations, the process may be shortened. There must
be as many given equations as there are unknown numbers.
4. Solve the equations
x + y\z= 9, (1)
x{y — z{'W=: 15, (2)
x — y\z]to = ll, (3)
Sx2y + 4:Z= 9. (4)
First eliminate w,
(2)(3), 2y2« = 4. (5)
Next eliminate x in (1) and (4).
(I)x3, 3a; + 3y + 3« = 27; (6)
(6) (4), &y^z=lS, (7)
(6)2, yz = 2; (8)
(7)(8), 4y = 16. ... y = 4.
Substituting 4 for y in (8), z = 2.
Substituting values of y and 2; in (1), x = 3.
Substituting values of x, y, and 2; in (2), t(7 = 10.
Solve the equations
3aj + 5y — 2 = 18, r7a; + 3y — 52 = 3,
5. <{5a; — 6.yf52;= 8, 8. ■] 3z — y — 4:X = l,
054^ + 2=6. ( X + y — Z=:l.
4a; + 22/42 = 22, r Sx3y \4:Z = 25,
6. ■{5x — 4:y\2z = 2Sy 9. } 5x^2y — z= 9,
5zx + Sy==2S. (4yllajf32; = ll.
2x + y\6z = A6, r 5aj — 4y4»=3,
7. ■{6x\^ySz = 16, 10. ] 3ajh6y22; = 14,
4cx'6y + 4:z = 12. ( 4.x + 5y — z = lS.
182
ALGEBRA.
[§306.
11. S6aj + 4y62; = 28,
6y — z + 7x = 56.
x — Sy'^2z= 0,
12. •^3x + 2y5z = 22,
x — y — z= 2.
13. 4x — y\z= 6,
(« — y — « = — 2.
i« + i3^ + i2 = 62,
5xSy __,
42j7a?" '
14.
15.
16. <
3z^x _^^
Sy^x
3^+_52 = i
42 + 5
17.
« — y — 2
yfx^z
= 8,
= 7,
18. <
19. <
y — « — 2
y — a?42 = — 2.
a; y 2 4
a
« z
1 1
20. <
7aj3y = l,
ll27v = l,
427y = l,
19 a?  3 V = 1.
PROBLEMS INVOLVING SIMULTANEOUS EQUATIONS.
Note. In the statement of a problem there must be as many equa
tions as there are letters representing unknown numbers (§ 306).
1. The sum of two numbers is 343, and their difference is
49. What are the numbers ?
2. Divide the number 89 into two parts such that ^ of the
greater part will exceed ^ of the less by unity.
3. A boy, being asked his age and that of his sister, replied,
" If I were 3 years older, I would be 3 times as old as my
sister ; but, if she were 2 years older, she would be J as old as
I am." How old was each ?
§306.] SIMULTANEOUS EQUATIONS. 188
4. If A were a years older than he is, he would be m times
as old as £ ; but, if B were b years older than he is, he would
be  the age of A. How old is each ?
n
6. A man has $ 10,000 in two investments. From the first
he gets 6% interest; and from the second, 4% ; yet the second
investment yields twice as much income as the first. What is
the amount of each investment ?
6. In an alloy of silver and copper, ^ of the whole, and
42 ounces more, was silver ; while the copper was 8 ounces less
than ^ of the whole. How many ounces were there of each ?
7. In an alloy of nickel, copper, and silver, J^'of the whole,
plus 8 ounces, is nickel ; •§■ of the whole, plus 4 ounces, is cop
per ; and ^ of the whole is silver. How many ounces of each
metal in the alloy ?
8. A certain number expressed by two digits is equal to
4 times the sum of those digits; but, if 18 be added to the
number, the digits will be reversed. Find the number.
9. Said A to B, " If you give me $ 100, 1 shall then have
as much money as you." — " Nay," replied B, " give me $ 100,
and then I shall have 3 times as much as you." How much
money had each ?
10. If 10 lb. of tea, with 35 lb. of sugar, cost $ 11.30, and
12 lb. of tea with 25 lb. of sugar cost $ 12.20, what is the
price of each per pound ?
11. If a lb. of tea with b lb. of sugar cost m cents, but
c lb. of tea with d lb. of sugar cost n cents, what is the cost
of each per pound ?
12. If the numerator of a certain fraction be multiplied by
2, and its denominator increased by 2, the result will be equal
to unity ; but if the denominator be multiplied by 2, and the
numerator increased by 3, the result will be equal to ^. What
is the fraction ?
184 ALGEBRA. [§ 306.
13. A certain fraction becomes equal to ^ when 3 is added
to both its terms, but becomes equal to ^ when the same num
ber is taken from both its terms. Find the fraction.
14. A certain fraction becomes equal to  if 2 be taken from
its numerator, and it becomes equal to unity if 3 be taken from
its denominator. Find the fraction.
15. A certain fraction which is equal to f is increased to f
by having the same number added to both its terms, and is
multiplied by 2 by having another number taken from both
its terms. Find the numbers.
16. A fishing rod consists of two parts. The length of the
upper part is to that of the lower part as 5 to 7 ; and 9 times
the upper part, with 13 times the lower part, is 36 inches more
than 11 times the length of the whole rod. Find the length
of each part.
17. A dealer has two sorts of tea. By mixing them at the
rate of 3 lb. of the finer to 5 lb. of the coarser, he can sell
the mixture at 95 cents a pound; but, mixing at the rate of
1 lb. of the finer to 3 lb. of the coarser, he can sell the
mixture at 90 cents a pound. What is the price per pound of
each kind of tea ?
18. The difference of two numbers is 8, and twice the sum
of their reciprocals is equal to 3 times the difference of their
reciprocals. Find the numbers.
19. Seven years ago A's age was just 3 times that of B,
but 7 years hence A's age will be just double that of B.
What is the age of each?
20. A person, wishing to give 25 cents each to a certain
number of persons, found he had not enough money by 25
cents ; but he could give each 23 cents and have 1 cent left.
How many persons were there, and what sum had he ?
§306.] SIMULTANEOUS EQUATIONS. 185
21. A cistern can be filled in 6 hours by two pipes running
together, and the one pipe fills as much of the cistern in J of
an hour as the other does in 1 hour. In what time could liiey
separately fill the cistern ?
22. There are three numbers, such that the first together
with I of the second is equal to 19 ; ^ of the second with f of
the third is equal to 23 ; and ^ of the third with ^ of the first
is equal to the second. What are these numbers ?
23. A certain number has three digits whose sum is 15.
The digit in the units place is 3 times that in the hundreds
place ; and, if 396 be added to the number, the order of these
digits will be reversed in the result. What is the number ?
24. A and B together earn $ 50 in 8 days ; A and C together,
$ 69 in 12 days ; B and C together, $ 55 in 10 days. How
much can each earn in a day?
25. A jeweler sold three rings. The price of the first with
J. that of the second and third was $25; the price of the
second with ^ that of the first and third was $ 26 ; and the
price of the third with ^ that of the first and second was
$ 29. What was the price of each ?
26. A person finds he can buy with $ 31.10 either 10 bushels
of wheat, 12 of rye, and 9 of oats ; or 12 bushels of wheat,
6 of rye, and 13 of oats ; or 16 bushels of wheat, 10 of rye, and
2 of oats. What is the price of each grain per bushel ?
27. There are four numbers such that, by adding each to
twice the sum of the remaining three, we obtain 46, 43, 41,
and 38 respectively. What are the numbers ?
28. A man has two horses, and also a saddle worth $ 10.
If he puts the saddle on the first horse, his value will be double
that of the second horse; but, if he puts the saddle on the
second horse, his value will be f 13 less than that of the first.
What is the value of each horse ?
186 ALGEBRA. [§307.
CHAPTER XII.
INVOLUTION AND EVOLUTION.
POWERS.
807. The power of a number is the product obtained by
taking the number one or more times as a factor (§ 27). The
factor taken one or more times is called the base of the power.
308. The exponent of a number denotes the degree of its
power (§ 29). The exponent 2 denotes the second degree ; the
exponent 3, the third degree; and the exponent n, the nth
degree.
309. A number that is composed entirely of equal factors
is a perfect power; and a number that is not composed of
equal factors is an imperfect power.
310. Involution is the process of raising numbers to required
powers.
Since any power of a number is the continued product of the number
by itself, involution involves the principles of multiplication.
311. 25^=25x25; (Saf^SaxSaxSa, and (3a)» =
3a X Sa X Sa ••• ton factors. Hence
Any power of a number may be found by taking the num
ber as many times as a factor as there are ernes in the eonponerU
of the required power.
1. What is the fourth power of 4 ? Of 5 ? Of 10 ?
2. What is the third power of a ? Oix? Oiy?
§315.] INVOLUTIOX AND EVOLUTION. 187
312. (ay = a? xa^xa^ = a^^^ = a^^^ = a^ and (a^y =
a"^ X a"* X oT ••• to n factors = a"^* = a*". Hence
Any power of a number may be raised to any required power
by multiplying the exponent of the given power by the exponent
of the required power.
3. What is the cube of a* ? Of a?*? Ofy*?
4. What is the nth power of a* ? Of or'? Ofy*?
5. What is the nth power of a ? Of6»? Ofaf?
Monomials.
813. 123 = (3 X 4)» = 3» X 4«; (3a)» = 3« x a«; and (3a)» =
3* X a". Hence
Any power of a number is ^/le ^wodwc^ of each of the factors
of the number raised to the required power,
1. What is the third power of 2x5? Of 4x3?
2. What is the fourth power of ooj* ? Of a^b^x?
3. What is the fifth power of 2 ab^c ? Of 3 a^xf ?
314. It is thus seen that any power of a monomial may be
found by raising each of its factors to the required power,
316. /'2Y = «x^ = 5^^ = ^;; /^2Y=2x^x2 = ^; and
\b) b b b xb b^' \b) b b b 6»'
(  ) = 2 X  X  ••• to w factors = — • Hence
\b) b b b 6»
A fraction is raised to any power by raising both its terms
to the required power.
4. What is the fourth power of ? Of ^?
b ar
6. What is the wth power of ? Of — ?
7 3r
188 ALGEBRA. [§ 316.
316. (fa)'=(+a)x(4a)x(+a)=ha'; and (+a)»= +a*.
(«)«=( a) x(a)=a^ ( a)8=(+a»)x( a) =  a»;
(— aF) X (— a)= 4a*; (+ a*) x (— a)= — a*; and so on. Hence
I. If a number is positive, all its powers are positive.
II. If a number is negative, its even powers are positive,
and its odd powers negative.
317. This principle may be expressed by the formulas
(a)" = a*; (a)*" = +a"*; and ( a)*"+^ =  a*»+i
318. It is thus seen that all even powers of a number are
positive, and all odd powers have the sign of the number itself
It follows that —a\ —a*, etc., are not perfect powers. See ** Imagi
nary Numbers'' (§ 413). The term number in §§316, 318, denotes a
real number ; i.e., a number that is not imaginary.
6. What is the third power of  3 d'h ? Of 4 ab^ ?
7. What is the third power of ^? Of ^? Of ^?
2b 3x 3 aoi
Raise to the indicated power
8. (Sa^by. 17. (' 2 a^icT^y. ^^ f 5xy \*^
9. (5aV)' IS. (3 abx)"^. \^^J
10. (^4.ab'xy. 19. (^^xyz)^, ^5. (^^Y.
11. (a^bcx^\ 20. (^abc)^^\ ^ "^ ^
12. (3a35^c)«. ^^ 26. (^Y
13. {2c^xf)\ ^ ^ ^ Va^y
15. (ahhir ^^ /zl2^^ 28. ( "T"'
16. (arb'^cy. \ 5ab^ J \ «/
819. To raise a monomial to any power,
Raise the numerical coefficient to the required power, multiply
the exponent of each literal factor by the exponent which indicates
the required power, and prefix the proper sign.
§322.] INVOLUTION AND EVOLUTION. 189
320. A fraction is raised to any power by raising both terms
to the required power, and prefixing the proper sign.
Binomials.
321. It may be found by actual multiplication that
(a + &)' = a^ + 2a6 + ^.
(a [ 6)3 = «» h 3 a^h + 3 a&2 + &^
(a + ft) * = a' + 4 a^ft + 6 a^ft^ + 4 a5» + b\
(a^bf = a'\ba'b^ 10a%^ + 10 a%8 + 6 a5* + ft*,
{abf = a^2ab^b\
(a  bf = o^ Sa^b {Sab^ b^
(a  by = a* 4:a^b + Ga^ft^ _ 4a6» + ft*.
(a ft)* = a'^  5a*ft + lOa^ft^  10 a^ft^ + 5 aft*  6».
322. An inspection of the above powers of a + ft and a — ft
shows the following facts or laws :
I. The number of terms in each power of the binomial is
one more than the exponent of the power.
II. The exponent of a in the first term is the same as the
exponent of the power of the binomial, and it decreases by 1
in each succeeding term.
III. The exponent of ft in the second term is 1, and it increases
by 1 in each succeeding term.
b does DOt appear in the first term, nor a in the last term.
The sum of the exponents of a and b in each term is the exponent of
the power of the binomial.
IV. The coefficient of the first term is 1, and the coefficient
of the second term is the same as the exponent of the power
of the binomial ; and generally the coefficient of any term is
the product of the coefficient of the preceding term by the
exponent of a in that term, divided by the number of that
term.
190 ALGEBRA. [§ 323.
For example, in the fourth power of a{b, the coefficient of
the third term is 4 x 3 * 2 = 6.
The coefficients of the first, second, third, etc., terms from the left are
respectively equal to the coefficients of the first, second, third, etc., terms
from the right.
V. When both a and b are positive, all the terms of the
power are positive ; and when b is negative, the odd terms are
positive, and the even terms are negative.
323. It is to be noted that the above equations denoting the
powers of a + b and a—b are identities, and are true for
all values of a and 6.
324. It will be shown later (Chapter XXI.) that the fore
going laws hold good for any power of a binomial indicated by
a positive integral exponent. Hence, if n denotes any positive
integer,
/ . 7.\« « . n !». . w(n— 1) ^ 2,2 , n(n—l)(n—2) ^ ..^ ,
This identity is called the binomial formula.
325. A binomial may be expanded to any power indicated
by a positive integral exponent by the application of the above
formula.
1. Expand (x—yf.
What is the number of terms in the expansion ?
What is the exponent of x in the first term ? In the third
term ? In the last term but one ?
What is the exponent of y in the second term? In each
sequent term ?
What is the coefficient of the first term? Of the second
term ? Of each sequent term ?
How is the coefficient of the third term found? Of the
fourth term ?
What is the sign of each term ?
§327.] INVOLUTION AND EVOLUTION. 191
Expand in like manner
2. (abf. 5. (axy. 8. (a + 6)*
3. (mny. 6. (x + yf. 9. (1  xf.
4. (a\xy. 7. (cd)*. 10. {xiy.
Write the fourth term of
11. (a by. 12. (c + d)^. 18. (xyy.
Write the first three terms of
14. (a by. 15. (c+d)« 16. {xyy^.
Write the last three terms of
17. (mny. 18. (a + xy. 19. (a — by.
326. When one or both terms of a binomial contain mord
than one factor, as Sa—l^, the binomial may in like manner
be raised to any power.
20. What is the third power oi2iX? — Sy?
Let a = 2a?, and b = Sy.
Formula. (a  by = a? 3a^b ^Sal^V.
Substituting 2 a? for a, and 3 y for b,
(2a?Syy=(2x^^S(2x^\Sy) + S(2a?)(3yy(Syy
=:Safi36a^y + 54aY  272/3.
Note. This is in effect the same as the inclosing of each term con
taining more than one factor in a parenthesis, and treating it as a or b.
327. Since each term, the first and last excepted, of a
binomial power higher than the first power, is composed of
three factors, the expansion of such a binomial power as
(2a^ — Syy may be somewhat facilitated by writing these three
factors in a column, and then forming their product as below.
Formula. {aby=a^ Sa^b H3a5« V.
(2a^' 3 +3 (3yy
(2a^2 2a^
3y (3yy
(2x'3yy = Sa^36a^h5^asy27y^
ld!2 ALoeBttiL. [§ 328.
Expand as above
21. (3a2by. 23. fab^» 25. (2m ny.
22. (a«36c)». 24. (^^cV. 26 (^"^^J'
Polynomials.
328. Polynomials of three or four terms may in like man
ner be raised to any power.
1. Wliat is the third power of 2 a' h 5 — c ?
2 a2 + 6  c = 2a« + (6  c).
Let X = 2 a^ and y = b — c.
Formula, (x + yy = afi + Sx^ + 80^2 + ^8
Substituting, (2a2)» +3 +3 +(6c)«
(2a2)a 2a2
(6c) (6c)2
(2a2 4 6  c)8 = 8 06 + 12 a*6  12 a*c + 6a^b^  I2a^bc
+ 6a2c2+ 68  362c + 36c5«  c8.
2. Expand (a + b^c — d)\
+ 6 + c — df=(a+&) + (c — d).
Let a; = a + 6, and y = c — (?.
Formula. (x + yy = i)fi + Sx^ + 3x2/2 + y».
Substitutmg, (a + 6)8 +3 +3 + (c  d)«
(a + 6)2 (a + 6)
(c  d) (c  d)2
(a + 6 + c  d) 8 = a* + 3a26 + 3a62 + 68 + 3a2c + 606c
+ 362c  3a2d  6abdSb^d + 3ac2  6ac(l + 3a<P
+,3 6c2  66cd+ 36^2 + c8  3c2d + 3cd2  d8.
329. Since the square of a polynomial is composed of (1)
the square of each term, and (2) twice the product of each
term multiplied in succession by the terms which follow it
(§ 161), the square of a polynomial may be directly written.
Thus, (a + 6 f c)« = a» 4. 6^ 4. c2 + 2a(6 4 c) + 2 6 X c
= a* + Z^^ + c» + 2 aft + 2 ac + 2 6c.
§331.] INVOLUTION AND EVOLUTION. 19*3
The cube of a trinomial is composed of (1) the cube of each term ; (2)
three times the product of the square of each term multiplied by the sum
of the other two terms ; and (3) six times the product of the three terms.
Eaise to the indicated power
4. (6a6Va*«)3. ' [my ' \ 2ab^cJ
6. (Sm^nx^yy. \ SnyJ \ xfz J
7. {fxf)\ ^^ /2aW ^^ /^
8. {da?yzy. '\2mrt'j' ' \ard
Min,n\ m
24.
(f^)"
Expand by the binomial formula
15. (m + w)*. 20. (12^*.
16. {mn)\ 21. (af^3a»)^
17. (a«&)». 22. (^^zy. 25. (»^J
1~ i' 23. (2a + f\ 26. (o^^^
19. (laj)«. \^ 2y V «/
Write the first three and the last three terms of
27. (x\y)^. 28. {xay\ 29. {x + y)^.
Write the sixth sequent term of
30. {acf. 31. (lxy^ 32. (xVf.
ROOTS.
330. The root of a number is one of the equal factors which
multiplied together will produce the number (§ 34).
331. The second or square root of a number is one of its two
equal factors; the third or cube root, one of its three equal
factors ; and the nth root, one of its n equal factors. Hence
the nth root of a** is a ; i.e., the root of any power is its base.
white's alq. — 13
194 ALGEBRA. [§ 332.
332. The root of a number may be indicated by the radical
sign, y/. Thus, V25 or V25 denotes the square root of 2b \
\/64, the cube root of 64 ; and VaS, the nth root of db (§ 35).
The figure or letter written in the opening of the radical
sign is called the index of the root.
The root of a number may also be indicated by a fractional exponerU*
1 • 1 5
Thus, a^ denotes the square root of a ; a", the nth root of a ; and a**, the
nth root of the mth power of a (§871).
333. Evolution is the process of finding the required root
of a number.
Evolution is the inverse of involution. Involution finds the product
of equal factors, and evolution finds one of the equal factors of a product
Monomials.
834. Since (3 a^&s)^ = 3« a^^^ft^^^' (§ 319) = 27 aW,
■</27a%^ = v^ X a«36^3 ^ 3 ^2^3^
Hence any root of a positive monomial is found by taJdrig the
required root of its numerical coefficient, and then dividing the
exponent of each literal factor by the index of the required root
Write the indicated root of
1. ViSbcM. 2. y/W^^. 3. y/U^HM^.
335. Since any even power of ± a is positive (§ 316), any
even root of a positive number is } or — . Thus,
^1/I6^ = ±2a;.
Since (+ a)2 = f a^ and (— a)* = + a\ the square root of — a* is
neither + a nor — o. The square root of — a^ can only be indicated
(V— a^), and the same is true of any even root of a negative number.
Such an indicated root of a negative number is called unreal or imagi
nary (§ 413).
A number has as many roots as there are ones in the index of the root.
Thus, v^lOas* = ± 2 a;2 and i V— 4 x*. The first two roots are real, and
the second two roots (± V— 4 a;* = ± 2 ac^V— 1) are unreal. Only real
roots are given in this chapter.
§339.] INVOLUTION AND EVOLUTION. 195
336. Since any odd power of + a is positive, and any odd
power of — a negative, any odd root of a positive number is
positive, and any odd root of a negative number is negative.
337. It is thus seen that the sign of the even root of a posi
tive number is \ or —, and that the odd root of any number has
the same sign as the number itself,
338. Since/^?Y=5^, ^1^ = ^=.^. Hence the root of
\b) 6» \6» y^ b
an algebraic fraction is found by taking the required root of both
its terms, and prefixing the proper sign.
Write the indicated roots of
4. V25a26^. 13. >^32aiV.
6. V81 a'b'<^. 14. V^[^. ^^ ^ ^"y"
6. ^256^. 15. VS^^^». 21. .^7^1^^
7. vu^w^, 16. v^;^^«.
„^ g/ 32a^6^<>
256m^^«
8. ^27mV. „/,^ * 22. a/^^
\ 17. VS^V*. \ 2/^«"
, o 16 a^62 «&V*
10. v/125mWs^. ^^ \o^^xl* 2^' \^^'
25a^ ^ sT
11. </81^yV^' 3/ 216 a«a^ ^4 ^H^
12. y/  243 a'^c^. ' \ 125 6^2* ' \ a«6***
Square Eoots of Polynomials.
339. Since the square of a ± 6 is a* ± 2a6 + 6*, it follows
that any trinomial is a perfect square, if, when properly
arranged, its extreme terms are perfect squares, and its middle
term twice the product of the square roots of its extreme
terms (§ 186). Hence
The square root of a trinomial that is a perfect square may
be readily found by arranging the terms according to some letter,
and then connecting the square roots of the extreme terms with the
sign of the middle term.
196 ALGEBRA. [§ 340.
Find by inspection the square root of
1. a?* — 4icy + 4y*. 5. 4a* — 4a + 1.
2. 4aJ* + 4aj*2/ + 2^. 6. l\ey^ + 9f.
3. a^6a^f + 9j/*. 7. Oa^  2a?y + i2^.
4. 9a'12ay2 + 4y*. 8. i^^a^ + y*.
340. The second term of the square root of a trinomial
which is a perfect square may also be found by dividing the
middle term of the trinomial, properly arranged, by twice the
square root of its first term.
341. Since the square of any polynomial is composed of (1)
the square of each of its terms, and (2) twice the product of
each term multiplied in succession by the terms which follow
it (§ 329), the square root of a polynomial that is a perfect
square may be found by inspection, provided the given poly
nomial contains only two different powers of some particular
letter. Thus, the square root of a*  2 at + 2 ac + ^^ + 2 6c + c*
is a + 6 f c; and this may be verified by squaring a\h + c.
342. It may assist, in finding the required root, if the poly
nomial be first arranged according to the descending powers of
some particular letter. Thus, a* + 6* + c* f 2 6c f 2 ac + 2 aft,
when arranged according to the descending powers of a,
becomes
a* \2a{h^ c)h V + 26(c)+ <?.
Find by inspection the square root of
9. «* + y*f42iry — 4a; — 4y.
10. a^\4:Qhj^\'^a^z\^7?\4.y^^l2fz\l^y^ + 97? +
2Az\ 16.
343. Since (a* + 2 a6 + 3 by= (a^ + 2aX2ab \Sb^{4t a^h^
+ 2(2a6 X 3 62) + 96*= a* + 4a«6 + lOa^t* + 12a6» + 96*, the
square root of a* + 4 a^6 + 10 a*62 + 12 a6^ h 9 6* is found by
reversing the foregoing process, as is shown below.
§343.] INVOLUTION AND EVOLUTION: 197
11. What is the square root of
a* f 4a«6 + lOaV f 12ab^ + 96*?
aH4a«6+10a262+12a68+96*  oH2fl6+36g , sq. root,
a*
'20^41 +4a«6+10a262 (1)
(2a2+2a6)2a6== +4af6f_4flW ^
2a2+ 6a2ft2_j.12a6S4.954 (2)
(2aH4q5+362)3&2 = 6a26H12a6«l96*
It is seen that the first term of the root (a^ is the square
root of the first term of the polynomial ; that the second term
of the root {\2ab) is obtained by dividing the first term of
the first remainder (1) by twice the first term of the root, and
the third term (3 b^ by dividing the first term of the second
remainder (2) by twice the first term of the root.
Note. If preferred, twice the first term of the root (2a^) may be
considered the first " trial divisor ;" and twice the first two terms of the
root (2a2 + 4a6), the second ** trial divisor."
The first subtrahend (a*) is the square of the first term of
the root ; the second subtrahend is twice the first term, plus the
second term, multiplied by the second term [(2 a* J 2 ab) 2 db'] ;
and the third subtrahend is twice the first two terms of
the root, plus the third term, multiplied by the third term
[(2a2 + 4a6 + 36^362].
For verification, square the root found by the binomial
formula, or, if preferred, by § 329, rearranging the terms of
the result, if necessary, and reducing to the simplest form.
Find the square root of
12. aj* — 4a^y + 6ic^2/^ — 4ir2/'4y*.
13. 9a^12a^y ^34x^^20x1^ + 251/*.
14. 50 a + 15a^ + 25 f a* lOa^.
15. l10a: + 27aj2_10a^4aj*.
16. 4ar* + aV4i2/'4ar'3^ + 2aJ2/aJ2^.
198 ALGEBRA. [§ 344.
17. l~4aj46y + 4aj*12ajy+92^.
18. 81  18a + a' + 186 2a6 + 61
19. ^a^ + ^a^ + ^a^^^f + y^^.
20. ^2xy + ^. 21. a^ + 3^+4«2a^ + ^^
22. l + 2x + 2y{2z + it^ + 2xy + 2xz^f + 2yZ'^i?.
23. l2a;f 3a^4aj»45a?*4iB»43a^2aj=^ + iB«.
24. a«6a*+15a«20 + ^4 + 4
a* . a* a"
344. To extract the square root of a polynomial,
Arrange the terms according to the powers of some letter.
Find the square root of the first term, and write it as the first
term of the required root, and subtract its square from the first
term of the polynomial
Divide the first of the remaining terms by twice the first term
of the root, and write the quotient as the second term of the root
Multiply twice the first term of the root plus the second term
by the second term, and subtract the product from the remaining
terms; and so on until no term remains.
Square Roots op Numbers expressed by Figures.
345. The smallest integer expressed by one figure is 1, and
the greatest is 9 ; the smallest integer expressed by two figures
is 10, and the greatest is 99 ; and so on.
346. The squares of the smallest and the greatest integers
expressed by one, two, and three figures, are as follows :
1*= 1 102= 100 1002= iQooo
9« = 81 W = 9801 9992 = 998001
It is thus seen that the square of an integer contains twice
as many 07'ders as the integer, or twice as tnany orders less one.
§ 350.] INVOLUTION AND EVOLUTION. 199
347. The squares of the smallest and the greatest numbers
composed wholly of units, or tens, or hundreds, are as follows :
12= 1 10*= 100 100*= 10000
9* = 81 90* = 8100 900* = 810000
It is thus seen that the square of units is units, or units and
tens; the square of tens is hundreds, or hundreds and thou
sands; the square of hundreds is tenthousands, or tenthou
sands and hundredthousands ; and so on.
348. It follows, that if an integer be separated into periods
of two orders each, beginning at the right, there will be as
many orders in the root as there are periods in the integer;
and hence the square root of the lefiJiand period of the integer
is the lefthand order of its square root
1. How many orders in the square root of 64? Of 625?
Of 1444 ? Of 273529 ? Of 45796 ?
2. What orders in 273526 contain the square of the units
of its square root ? The square of the tens ? The square
of the hundreds ? What is the first or lefthand figure of the
root ?
3. How many orders in the square root of 145796 ? What
is the first figure of the root ?
349. An integer may be separated into periods of two
figures each by beginning at units, and placing a dot over
each alternate figure, thus, 145796; or by placing the dot
between the periods, thus, 14*57*96.
Separate into periods and give the first figure of the square
root of
4. 626. 6. 94249. 8. 5306845.
6. 3026. 7. 492804. 9. 54756090.
350. The square of an integer may be found by the binomial
formula, as shown below.
802 = 6400
2x80x5= 800
62= 25
200 ALGEBRA. [§ 351.
10. What is the square of 85 ?
85 = 80 + 5.
Let a = 80, and 6 = 5.
Since (a + 6)2 = a^ + 2 a6 + h\
(80 + 5)2 = 802 + 2 X 80 X 5 + 52 = 7225
It is thus seen that the square of any number composed
of tens and units equals (1) the square of the tefiis, (2) plus
twice the product of the tens multiplied by the units, (3) plu^ the
square of the units.
351. Since the square of tens gives no order lower than
hundreds, and the product of the tens by units gives no order
lower than tens, the tens and units of the root may be found
as shown below.
11. What is the square root of 7226 ?
. . tu
7225185
a2 = 82 = 64_
2a = 2x80 = 160)825
(2a + 6) 6 =(160 + 5) x 5 = 825
The same result may be obtained by omitting the unit figure in l>oth
the trial divisor (160) and the remainder (825): thus,
. . tu
7225186
82= 64
Trial divisor, 16)825
165 X 5 = 826
12. What is the square root of 104976 ?
. . . htu
f^ 10 49 76 1324
(a = 3) 32 = 9
Trial divisor, 2x3 = 6)149
^(20x3 + 2)2 = 62 x2= 12 4
(a = 32) Trial divisor, 32 x 2 = 64)2 576
(320 X 2 + 4) 4 = 644 X 4 = 2 57 6
§354.] INVOLUTION AND EVOLUTION. 201
It must be observed that a represents first 3 hundreds, considered as
3 tens with respect to the next figure of the root ; and that, in finding the
third figure of the root, a represents 32 tens, the part of the root already
found.
352. Since .4* = .16, .04* = .0016, etc., the square of a deci
mal has twice as many decimal orders as the decimal; and
hence the square root of a decimal has one half as many
decimal orders as the decimal. A decimal is separated into
periods by beginning at the decimal point, and pointing off to
the right ; thus, .0625. A decimal cipher must be added if the
decimal contains an odd number of orders ; thus, .6260.
13. What is the square root of 13.3225 ?
13.3225 [3.65
32= _Q
3x2= 6)4.32
6.6 X 6 = 3.96
3.6 X 2 = 7.2). 3625
7.25 X .05 = .3625
When an integer or a decimal is not a perfect square, its root may
be found approximately by adding periods of decimal ciphers. Thus,
V32 = V32.0000 = 5.65+.
353. The square root of a common fraction is found by
extracting the square root of each of its terms. Thus, VJ = f .
When the denominator of a common fraction is not a perfect
square, its square root can be found approximately by multi
plying both of its terms by the denominator, and then extract
ing the square root of both terms of the resulting fraction,
carrying the root of the numerator to two or more decimal
places. Thus,^=V^=^=^=.75+.
A common fraction may also be changed to a decimal, and the square
root of the decimal found.
354. When the righthand period of a decimal contains
only one order, a decimal cipher should be annexed, 4.322
thus becoming 4.3220,
202 ALGEBRA. [§ 355.
Find, the square root of
14. 69169. 21. 176.89. 28. f.
15. 94249. 22. 45.1584. 29. 272^.
16. 57600. 23. .008836. 30. 1040^^.
17. 210681. 24. .000625. 31. ^.
18. 492804. 25. 75.364. 32. ^.
19. 522729. 26. 586.7. 33. m.
20. 390625. 27. .056644. 34. ^V^
Find to three decimal places the square root of
35. 2. 38. 3.5. 41. f. 44. f
36. 3. 39. Q,^, 42. f. 45. ^.
37. 5. 40. 0.9. 43. f 46. ^.
Cube Koots op Polynomials.
355. The process of extracting the cube root of a polyno
mial is readily derived from the formula
(a ± 6)8= a8 ± 3a% ^^aV ± h\
This identity shows that a ± 6 is the cube root of
It is observed that Va^ or a, is the first term of the cube
root ; and ± 3 a^6 ^ 3 a^, or b, the second term of the root,
3a^ being the trial divisor; and since
±^a?h + 3a62 ± h^ = {Zd? + Sdb+hy),
3 a^ + 3 a6 + 6^ is a complete divisor of the last three terms of
the polynomial, h being the other factor.
356. It is thus seen that the cube root of c?\Z a*6+3 a6*f6'
may be found by taking the cube root of a^ for the first term
of the root, and dividing 3 d?h by 3 a? for the second term of
the root. Thus,
§357.] INVOLUTION AND EVOLUTION. 203
(af = aP_
Trial divisor, So" Sa^ft+Saft^+fts
Complete divisor, 3a^+3a6f ^^
(3d'+Sa^+b^)b=S^b±Sa^±^
1. What is the cube root oi a^ + 9 a^y { 27 ith/^ + 27 f?
«6+9iK*yH27a2y2+27 y8[a;f+3^
a* (a;2)8= ^6
3a2(T. D.) ... 3aj*
3a2+3a&+62(C.I>.) Sx^{9x^{9y^
9x*y+27xV+27y«
(3a^ + 9g2y49y233y:^ 9a^y4.27g2y2._27y8
357. A cube root having any number of terms may be found
in like manner if it be observed that each successive trial
divisor is three times the square of the part of the root
already found ; and that each successive divisor is completed
by adding to the trial divisor (1) three times the product of
the root term last found multiplied by the part of the root
before found, and (2) the square of the term kist found.
2. What is the cube root of
8a^ 36ic* + 66 a?*  63a^ + 33a^  9aj + 1?
I2a;23a;+l
8ir«36iB6+66aj*63x8+33x2_9a;__i
(2a;2)8= 8^
(T. D.) 3(2a;2)2= 12a^  36a*+66rK*63a^
(12«*18a;8+9«2)(_3«)= 36a^+64a^27a^
(T.D.)3(2x23x)2= 12a^3 6a;8427a;2  12a:*36«8+33a;29a;+l
(12x436«8+27xH6a;29x+l)l= 12a:*36a^433g29a;+i
Find the cube root of
3. iB« + 6aa^ + 12aV+8a«.
4. 8a^12aar* + 6a2ajal
6. 125 a«  225 a%^ + 135 a%*  27 b\
6. a«43a* + 6a* + 7a3 + 6a2 + 3a + l.
7. aj« + 6a^40iB8 + 96aj64.
204 ALGEBRA. [§ 358.
8. 8aj«12aj«54aJ* + 59aj«f 135ar^75aj125.
9. l+9» + 18a^27aj354aJ*f 81a?527i««.
10. aj86aj + — ,•
X or
11. a83a* + 9a13 + — ^ + 
a (T a^
12. 27 a«108 a'b+90 a'b^+SO a^ft^BO a26*48 a6*~8 b^
13. 8 a:^  36 ar'^/ + 66 xY  63 xY + 3Sa^y^ 9xf + f.
14. 27 a^  54 aar' + 63 aV  44 aW + 21 aV  6 a^a? + a^.
358. To extract the cube root of a polynomial,
Arrange the terms according to the powers of some letter.
Find the cube root of the first term, write it as the first term
of the root J and subtract its cube from the polynomial
Divide the first of the remaining terms by three times the sqvxire
of the first term of the root (trial divisor), and write the qriotient
as the second term of the root.
To the trial divisor add three times the product of the fifst
term multiplied by the second, and the square of the second; and
then multiply the complete divisor thus formed by the second term
of the root, and subtract the product from the remaining terms
of the polynomial.
Proceed in like manner until all the term^ of the polynomial
are used.
Cube Eoots of Numbers expressed by Figures.
359. The cubes of the smallest and the greatest integers
expressed by one, two, and three figures, are as follows :
13= 1 103= 1000 1003== 1000000
9^ = 729 99^^ = 970299 999^ = 997002999
It is thus seen that the cube of an integer expressed by one
figure contains from one to three orders; that the cube of an
integer expressed by two figures contains from four to six
orders; and that the cube of an integer expressed by three
figures contains from seven to nine orders.
§364.] INVOLUTION AND EVOLUTION. 205
360. It follows, that the cube of an integer contain^ three
times as many orders as the integer, or three times as many
orders less one or less two,
361. The cubes of the smallest and the greatest numbers
composed wholly of units, or tens, or hundreds, are as follows :
13= 1 103= 1000 1003= 1000000
^ = 729 903 ^ 729000 900^ = 729000000
It is thus seen, that if a number be separated into periods
of three orders each, beginning at the right, the first period
will contain the cube of the units of its cube root ; the second
period, the cube of the tens of its cube root ; the third period,
the cube of the hundreds of its cube root ; and so on.
362. It follows, that the cube root of an integer contains as
many orders as there are periods in the integer, and the cube
root of the lefthand period of an integer is the lefthand term of
its cube root,
363. An integer may be separated into periods of three
figures each by placing a dot over the first or units, fourth,
seventh, etc., orders, thus, 48228544; or by placing a dot
between the periods, thus, 48*228*544.
Separate into periods and give the first figure of the cube
root of
1. 42875. 3. 117649. 5. 274625.
2. 91125. 4. 185193. 6. 9405424.
364. The cube of a number may be found by the binomial
formula (a ±lif = a^ ±3a% \^ab^ ±W, d^ shown below.
7. What is the cube of 85 ?
808 = 512000
85 = 80 + 6.
Let a = 80, and & = 5.
Since (a + &)« = a^ + 3 a^ft \Sab^+ h^
3 X 802 X 5 = 96000
3 X 80 X 52 = 6000
58 = 125
(80 + 6)8 = 808 + 3 X 802 X 5 f 3 X 80 X 52 + 68 = 614125
206
ALGEBRA.
[§ 365.
365. It is thus seen that the cube of any number composed
of tens and units is equal to (1) the cube of the tens, (2) plus
three times the product of the square of the tens multiplied by
the units, (3) plus three times the product of the tens multiplied
by the square of the units, (4) plus the cube of the units.
Find thus the cube of
8. 82. 10. 77.
9. 68. 11.. 66.
12. 104.
13. 125.
14. What is the cube root of 262144 ?
«»+ (3a2 + .3a6 + &2)6
a*
(T. D.)
3a2
68 =
. . . 3 X 602 = 10800
3 a6 . 3 X 60 X 4 = 720
62 . . 42= 16
t u
262144 64
216
46144
(3a2 + 3a6 + 62)5
= 11536 X 4 = 46144
It is seen that the first trial divisor is 3 x 6O2 = 10800, and that the
complete divisor is 11536.
15. What is the cube root of 16387064 ?
a«
(2)8 =
16387064 [264
8
3a2 3 X 202 = 1200 8387
Sab . . 3x20x5= 300
62 62 = 25
(3a2 43a6 + 62)6
3a2 3 X 2602
3a6 . . 3 x260 x4
62 42
(3a2 + 3a6 + 62)6 = 190516 x 4 = 762064
1525 X 5 = 7625
187500
3000
16
762064
It should be noted, that, in finding the first two terms of the root
(25), a in the formula denotes hundreds, and 6 tens (hundreds being
considered tens, and tens units); but, in finding the third term (4),
a denotes 25 tens, and 6 the units sought.
§ 867.] INVOLUTION^ AND EVOLUTION. 207
The process may be continued to any number of terms in the root
by observing, when finding a new term, that a in the formula denotes
the part of the root already found.
366. Since .!« = .001, .Ol^ = .000001, etc., the cube of a deci
mal has three times as many decimal orders as the decimal ;
and hence a decimal is separated into periods of three orders
each by beginning at the decimal point, and pointing off to the
right, thus, .015625. . If the last period does not contain
three figures, decimal ciphers must be added ; thus, .262500.
367. Since the cube of a fraction is obtained by raising each
term to the third power (§ 315), the cube root of a fraction is
found by eoctracting the cube root of each of its terms. If the
denominator of a fraction is not a perfect cube, its approxi
mate cube root may be most readily found by first changing
the fraction to a decimal.
Find the cube root of
16. 42875. 23. 97.336. 30. 5fjf.
17. 185193. 24. .097336. 31. 37^.
18. 3048625. 25. 1953.125. 32. ^^
19. 48228544. 26. 67.419143.
20. 34328125. 27. 28.094464. ^3.
1.728
1.5
.216
21. 27270901. 28. ^VA « ^
22. 74.088. 29. 1444. .0625
Find to three orders the cube root of
35. i^. 38. g. 41. ■^^.
36. y. 39. g. 4^. YTj".
37. 10. 40. i. 43. ^.
208 ALGEBRA. [§ 368.
368. To extract the cube root of a number,
Begin at units, and separate the number into periods of three
figures each.
Find the greatest cube in the lefthand period, and write its
cube root as the first term of the required root.
Subtract the cube of the first term of the root from the left
hand period, and to the difference annex the second period for
a dividend.
Divide this dividend by three times the square of the first term
of the rooty tvith ttvo ciphers annexed {trial divisor), and write
the quotient as the second term of the root.
To the trial divisor add three times the product of the first term
multiplied by the second, and the square of the second; and
then multiply the complete divisor thus formed by the second term
of the root, subtract the product from the dividend, and to the
difference annex the next period OjS another dividend.
Proceed in like manner until all the figures of the given numr
ber are used.
369. The fourth root of a number may be found by extract
ing the square root of its square root ; and the sixth root, by
extracting the cube root of its square root, or the square root
of the cube root. The cube root of the square root is prefer
able in practice. Thus,
\/G25=VV625=V25 = 5; \/4096 = ^^ V4096 = "^ = 4.
Find, as above, the fourth root of
44. 16 a* + 96 a^b + 216 a'b^ + 216 ab^ \ 81 b\
45. ic88a;^f16aJ«+16ar^56a;*32aj3+64a^ + 64aj + 16.
Find the sixth root of
46. a^ + 6a^2/f 15iC*2/^ + 20a^2/8 + 15ajyf ean/' + Z.
47. 1 + 12a + 60a2 + 160a3 + 240a* + 192a^ 4 64a«.
§ 374.] RADICALS. 209
CHAPTER XIII.
RADICALS.
370. A radical is the indicated root of a number ; as Voib.
The term radical is also applied to expressions that contain a radical ;
as 3 aVai), and 3 a + Vab (§§ 375, 394).
371. If the indicated root can be exactly obtained, the
radical is called rational; if the indicated root cannot be
exactly obtained, the radical is called irrational or a surd. Thus,
a/25 and Va^ are rational, and V5 and v^ajft are surds.
Roots may also be indicated by fractional exponents, as 6*, (a + 6)*
(§ 332) ; and roots thus indicated may be rational or irrational. Thus,
6^ and {a + hy are surds.
Irrational numbers are also called incommensurable^ since they have
no common measure with unity.
372. It is to be observed that algebraic surds may become
rational when numerical values are assigned to the letters. Thus,
if we make a = ^, and 6 = 3, the surd Vah h becomes rational.
Surds that can be rationalized are said to be surds in form,
373. The degree of a radical is indicated by the index of the
radical sign. Thus, VoS (index 2 understood) is a radical of the
sec(ynd degree ; Va f 6, a radical of the third degree ; and so on.
Radicals of the second degree are also called quadratic
radicals ; and those of the third degree, cubic radicals.
The degree of a radical is also called its order. Since surds are a class
of radicals, what is true of radicals generally is true of surds.
374. The coefficient of a radical is the factor placed before
the radical part. Thus, in the expression 6^ a — h, 5 is the
white's alo. — 14
210 ALGEBRA. [§ 375.
coefficient of Va — 6; and in the expression 7a^/bx, 7 a is
the coefficient of Vbx,
375. When a radical contains no coefficient (except 1 under
stood), it is said to be eivtire; and when it contains a coefficient,
it is said to be mixed. Thus, Voi and Va — a? are entire radi
cals, and 2V5 and 5 a Va? — y are mixed radicals.
The coefficient of a radical is called the rational factor ; and the radical
part, if a surd, the irrational factor.
376. Similar radicals are those which have the same index,
and the same number under the radical sign ; i.e., have their
radical parts identical. Thus, 2 V3 and aV3 are similar ; so,
also, are 3^—2 and 2a^— 2.
377. Since Va x \/h= Va6, and, conversely, Va6= Va x V&,
it follows (1) that the product of the same roots of two factors
eqvxds the same root of their prodiict; and conversely (2) that
the root of any product equals the product of the same roots of
its factors.
378. This principle enables us to reduce a mixed radical to
an entire radical, and an entire radical to a mixed radical ; also
to make other important reductions of radicals in degree and
in form, as shown below.
REDUCTION OF RADICALS TO EQUIVALENT RADICALS OF
DIFFERENT DEGREE.
379. Since a = Va^, or Vo?, or Va"*, it follows that any
rational number may be changed to an equivalent radical of
any degree by raising it to the power corresponding to the index
of the radical, and placing the result under the radical sign.
Thus, 005 = ^aV; 3~\gT; and a — 6 = V(a —
by
§ 383.]
RADICALS.
211
380. In like manner a radical of any degree may be changed
to an equivalent radical of a higher degree by mvltiplying the
index of the radical, and the exponent of each factor under the
radical sign, by the same number. Thus,
V5 = ^= ^5/125; ^/a^ = ^/a^', and Va + b = V(a + b)\
381. Conversely, any radical may be changed to an equiv
alent radical of a lower degree by dividing the index, and the
exponent of each factor under the radical sign, by the same
number. Thus,
■y/a^=s/^'^ </(a h 6) = VoT^ ; and v^o^ = Vo^.
Reduce the following radicals to equivalent radicals of the
indicated degree :
1. a6 to third.
2. oj — 2^ to second.
3. ^ to fifth.
or
4' Va» to sixth.
5' >/x\y to sixth.
6« V(a — xy to nth.
7« VaS to mth.
8 y/c^b^ to second.
^' ^(^ — yf to nth.
0. V^,.....
{^ + yy
to second.
1.  to fourth.
x
If a; —
to sixth.
x — y
3. "\/a*V*d^ to nth.
6)
3
to second.
382. Radicals of different degrees may be reduced to
equivalent radicals of any degree which is a multiple of their
indices. Thus, VB and y/l may each be changed to the sixth
degree, as above; and y/ac, Va — b, and Va 4 b may each
be changed to the twelfth degree.
383. This process is called the reduction of radicals to a
common index. When the common index is the L. C. M. of the
indices, it is a common index of the lowest degree.
212 ALGEBRA. [§ 384
Reduce to a common index of the lowest degree
16. V6, n, and ^10. ^^ & J2 ^^ ^
16. Var^, Vxy, and wa^j^. r r~ r~
17. ^, .5/26, and ^^«. ^^^ \a' \h "^^ \?
20. a?, y/och/f and VaJ^^.
21. y/x — y and ^/x + y,
22. a — 6, ^(a — 6)*, and v/(a — &)^.
23. V^ft, lj^, ^I^.
+ 6y
REDUCTION OF RADICALS TO SIMPLEST FORM.
384. A radical is in its simplest form when its radical part
is integral, and contains no factor which is a perfect power of
the same degree as the radical. Thus, Va^ — 6* and 3Va^
are radicals in their simplest form.
385. When the number under the radical sign contains a
factor which is a perfect power of the same degree as the
radical, the radical may be reduced to its simplest form by
removing such factor from under the radical sign, and making its
proper root the coefficient, or a factor of the coefficient. Thus,
■\/25 a^x = V25 a^ x x = \/25 a^ x Va = 5av^;
3^/1^ = 3^/^ X \/P = 3a\/P.
Reduce to simplest form
1. V75. 4. fv^l92.
2. v'320. 6. V32^*.
3. 2</80. 6. 3a^6256y.
§387.] RADICALS. 213
7. V125 a»  50 a*6. 10. V(x y)(x''f),
8. ^16a5y24aY H V^C^ + y)'(«'yO.
9. V(a4&)(a^2>^ 12. (« + y)VaJ^ 2ic*y + ary^
13. (a 4 5) V3a2&2  3006^ 4 75 6*.
14. 5 a Va? — 2 ajy + o^.
15. (a + y)Va* — icV
386. When a radical is fractional, it may be reduced to its
simplest form by multiplying both terms of the fraction by such a
number as wiU make the denominator a perfect power of the
degree of the radical, and then proceeding as above. Thus,
Reduce to simplest form
16. V^. 18. ^. 20. 10 Vf a^x,
17. ^. 19. 3^. 21. lOa^fe^— ?.
^ 25 aW
23. iJ ^^'. 26. !c'«'\/ ^
24. 6aJ^ni. 27. £z:lj/27(^ + l).
\3(o6) 3 >( (a;l)»
387. When the quantity under the radical sign is a perfect
power whose exponent is a factor of the index, a radical may
be simplified by dividing both the index and the exponent by the
common factor. Thus,
■V^='^/c^; ■^(a + W = Va + b; and ■^(aby=(aby.
214 ALGEBRA. [§ 388.
Eeduce to simplest form
28. </IOO. 32. ^(a + &)'(a  6)«.
29. V125aV. 33. V4 a*  24 a6 + 36 &*.
30. </(a'  by. 34. '^a^y^\xyy.
31. A/(a2  6*)Xa + i>)*. 35. Va^ft^a + 6)^.
INTRODUCTION OF COEFFICIENTS UNDER THE RADICAL
SIGN.
388. The coefficient of a radical may be placed under the
radical sign by raising it to the power corresponding to the index,
and introducing the residt under the sign as a factor. Thus,
6 Vai = V25 X VoS = ^250^]
5 a y/xy = ■y/25c? x y/xy = V25 a^xy.
It will be observed that this is the converse of § 385.
Thus, V4 a^oj = 2 a Vx, and, conversely, 2 a y/x = V4 a*a?.
Introduce under the radical sign the coefficient of
1. 2 v^oS. g m + 71 / m — n
2. 3aVi06. mn\m + n
3. V5^. 9. _J_^w3_^8.
4. V^.
2
m — w
^4aV
4a^ar^
6. 3ic '
13. («&)V^ —
^a — 6 a —
b
6
§ 390.] RADICALS. 215
ADDITION AND SUBTRACTION OF RADICALS.
389. Since the coefficient of a radical denotes the number
of times it is taken, it is evident that the sum or difference of
two similar radicals is found by prefixing the sum or difference
of their coefficients to the common radical part. Thus,
7Va6 + 5Va& = 12 Va6; and 7 Va— 6 — 5Va — 6 = 2^ a — 6.
390. If the radicals to be added or subtracted are dissimilar,
they must first be changed to similar radicals (§ 376). When
this cannot be done, their sum or difference can only be indi
cated by the proper sign.
1. Add 7V8, 3V32, and  SVJ
7V8 = 7Vr>r2= 14 V2
3 V32 = 3Vi6x^ = 12V5
 8Vi =  SVfx^ = 4 V2
Sum= 22v^
2. Simplify V8(a  h) f 4 Vl8(a  h)  12Vi(a b).
v'8 (a  6) = V4 X 2 (a  6) = 2V2(a6)
4 Vl8 (a  6) = 4 V9 x *2 (a  6) = 12V2(a6)
 12Vi(a6) =  12V32j(a6) =  3\/2 (a  6)
Value = llV2(a&)
If the given radicals are of different degrees, they most first be reduced
to the same degree (§§ 380, 381).
Simplify
3. Vi8 + V32. 8. 2</250 + 3</54.
4. V54 + 2V294.> 9. y^logo ~.>/iO +^135.
5. 12V203V45. 10. 3 V80 + 6 V45  2 Vi25.
6. 4V6Vp. 11. VJ2ViVS.
7. 16V3^4VVl. 12. V2+V8VJVi.
216 ALGEBRA. [§ 391.
13. 4Vi + iVi3V^. 20. Vl8VJ+Vi+3VS.
14. V\+V^+^^. 21. 3ay/^5aVxy*
15. </48+</^ + 8</^ 22. 5a'V^5axV^.
16. ^2705^80+^640. ^^ 3a»<^3a^^^.
^a >a^ ^ar
25. 3^c  5v/? + 2^.
26. 5 V(a — xf — 2 axVa — a?.
 444
391. To find the sum or difference of radicals,
Reduce the radiccUs, if necessary, to similar radicals, and
then prefix the sum or difference of the coefficients to the radical
part.
If the radicals cannot he made similar, indicale their sum
or difference by the appropriate sign,
MULTIPLICATION OF RADICALS.
392. Since a^/a x b^=a xb x \/a x V& = ab^s/ab, it fol
lows that the product of two or more radicals of the same
degree equals the product of their coefficients prefixed to the
product of their radical factors.
If the radicals are of different degrees, they must first be
reduced to equivalent radicals of the same degree.
1. What is the product of 6V8 and 7V6?
6V8 x7\/6 = 6x7x VSx^ = 42\/l6~x3 = 168V3.
2. What is the product of 3 V2 and 5^4 ?
.. 3V2 X6\^ = 3\^8x6v^i6 = 3x6\/8^ri6 = 16v^6rx2 = 30v^.
§394.] RADICALS. il7
3. What is the product of Va and y/h?
Simplify
4. 6V8x7V2. 18. 3Vaax2V5^.
5. 5vT0x2vl5. 19. 3V^ X 2V5a» X 3Va.
6. ^^v/12X^.
 ^g X ^>§
8. 2Vx3Vf ,^. 2aM^3JIg.
9. VxA/fxV. ^52 >i8a!V
10. 4v^ X 5\/9 X 2\/4. 22. 3aV^x2bVa\
11. 12\^ X 2<^ X 3^^.
8^
23. 3\P^ X 4v/a  «.
^a — oj
12. VlBxViO.
13. 2 V3 X 3v^ X ^. 24. aVS x 6^? x c^.
14. 2^3x2V4x3V2. 25. V3^x ^5/4^x^2^^.
16. 2vi X 3v^ X V. ^aj2 — 2^
17. 4Vo^x3</^. 27. a/5 X \/^^^25 X ^/iO.
393. To find the product of two or more radicals,
Reduce the given radicals, if necessary, to equivalent radicals
of the same degree; then prefix the product of the coefficients to
the product of the radical factors, and reduce the resulting radical
to its simplest form.
394. A compound radical is a polynomial that has one or
more radical terms; as, 2 Va + 3V6, a — V3^, Va — Va — &,
and \Ja\ 6— Va— 6
218 ALGEBRA. [§ 395.
28. Multiply x + 2V3 by a; VS.
X +2V3
«  V3
x2 + 2 X VS
 gV32VP
The product of two binomial radicals may be written directly as
in § 166.
Multiply
29. V5 — 3 by Va 4 2.
30. 2 Va  4 V3 by 3 V« + 2 V3.
31. 2V5 + 3V2 by 3V5  4V2.
32. 2Va3V6 by VaHA/6.
33. 4Va+V6 by 4Va + V6.
34. a: + V^ + 2/ by V« — V^.
35. aHaV3Hl by aV3 — 1.
36. \/3 + 2\/2 by 2\/3 + ^.
37. 2V + 3Vi + l by 3Vi2VJ.
395. When two binomials involving radicals differ only in
the sign of a radical term, they are said to be conjugate ; and
their sum and product are rational. Thus,
(a + V6)(aV6)=a26, and (VaA/&)(Va+V6)=a6.
Multiply
38. 2^—^ by 2Va + V6.
39. yfx + 2 V^ by V5 — 2V^.
40. 3 V5 + 2 V3 by 3V5  2V3.
41. ^v^Vl2 by ^VS+Vi2.
42. Va;V21 by V^TV^I.
43. l+Vaj + l by lVa + l.
44. Va+^— Vo^^ by Va 4 a; 4 Va — ox
397.] RADICALa 219
DIVISION OF RADICALS.
396. Since ay/x x hVy = db^^/xy, db y/xy j hVy = aVx ;
and hence the quotient of two radicals of the same degree
equals the quotient of their coefficients prefixed to the quotient
of their radical factors.
If the radicals are of different degrees, they must be reduced
to radicals of the same degree before dividing.
1. Divide 12V75 by 5V3.
12 V76 ^ 6\/3 = ^VJ^ = V^^ = 12.
2. Divide V^ by >^.
Divide
8. 6Vl08 by 3V6. 12. V3f by ^.
4. 3V8 by 6V27. 13. \/2^ by V2}.
6. 3V6 by V8. 14. 2V5 by ^5.
6. 3vl^ by 2V. 15. 15V^ by ^Va^.
7. 12 Vf by 3VJ. 16. 2aV^ by 5aj^v^^.
8. ^A^ by ^^. 17. Sa^v^'^ by 10 xV^.
9. 6V86V2 by 3V2. is. 5V? by 4\/?.
10. Vl by ^6. 19. 3aVa«5 by 2a\/cM.
11. ^^I2 by Vs. 20. Va^y* by V«^.
397. To find the quotient of two radicals,
Reduce the given radicals, if necessary, to equivalent radicals
of the same degree; then prefix the quotient of the coefficients
to the quotient of the raduxU factors^ and reduce the remiUing
radical to its simplest form.
220 ALGEBRA. [§ 398.
BATIONAlilZINO THE DlYISOB.
898. If the denominator of a fraction is a radical, the denom
inator may be made rational by multiplying both terms of the
fraction by such a number as wiU make the denominator a per
fect power of the degree of the radical.
Thus, X=AxVL=?V3^^.
V3 V3xV3 V9
2 X Va^ '^^^^ '^^^^
a' :5 V a
^/a ^/ax^/c^ ^/^ ^
Reduce to equivalent fractions with rational denominators
1. A, 6. ^^. 9. '
V3 V6 </S^'
2. 2/1. « VlO 10. ^
V3 2Vfi 2V6
u>
V6
6.
Vio
2V6
7.
3
8.
5
3. i5.. ^ 3 11. "
V2 " m </f
4. ^. 8. ^. 12
Vl2 V3s^
■4
399. In like manner the quotient of two radicals may be
found by rationalizing the divisor, and then reducing the quo
tient to its simplest form. Thus,
400. When a divisor is a binomial of the form Va ± V&, it
may be rationalized by multiplying both dividend and divisor
by the conjugate of the binomial divisor (§ 395). Thus,
V3 ^ V3x(V3+V2) ^ 3+V6 ^3^^/g
V3V2 (V3V2)(V3fV2) 32
§ 401.] RADICALS. 221
Eationalize the divisor and simplify
13. V8l^V3. 23 V8
14. Vios^Ve. V5IV3
3+V5
15. V48fVi2. 24.
16. 3V85V6.
17. Vi5^2V5. ^^•
3V5
V3+V2
V3V2
18. (V3V2)V6. 2^ V^+V6
19. V3^Vl0. VaV6
6 27. ^V^ + ^V^ 
' 3V32V2'
V7
5ViO + 2V3
2V3f 3V2
3V3 44V2
A*vr.
6V7
21
1
ax.
Vio3
29
21
40^.
Vio+V3
L _
1
28.
29.
30. 31. ' 32. .
IV24V3 1+V2V3 2+V3+V5
Suggestion. In Example 30, multiply both terms by (1 — V2) — V3,
and then multiply both terms of resulting fraction (simplified) by — 2\/2.
401. The above method of rationalizing the divisor has
special value in finding the approximate numerical value of a
quotient when the divisor is a surd^ as shown below.
33. Find the approximate value of 4 « VS.
J_^4V3^V38==?:M± = 2.309+.
V3 3 3 3
5
34. Find the approximate value of
4V2
6 ^ 6C4 + \/2)  2Q + ^^  2Q + ^^^  1.0336 1,
4V2 (4>/2)(4 + >/2) 1^2 14
222 ALGEBRA. [§ 402.
Find to three decimal places the value of
36.
2
V8
86.
10
V5
37.
6
Vl2
3ft.
6
<IQ
3
Oo»
V22
AO
V3
t\9*
2tV3
41
2V3
tbx.
3V2
iiO
V5V3
d^
6
%o.
V3+V2
44
V6+V2
w.
V6
dfi
3+V5
%o.
3V6*
46.
V8
V27 V5 + V3 V8V2
INVOLUTION AND EVOLUTION OF RADICALS.
402. Radicals may be raised to any power by substituting
fractional exponents for the radical signs, and then proceeding
as in the involution of numbers with positive integral expo
nents (§ 319). The roots of radicals may be found in like
manner. For other operations with fractional exponents see
Chapter XIV.
1. Find the square of 4Va.
(4v^)2 = (4 a*)2 = 16 a* = 16 v^.
2. Find the cube of 3 Vod*.
(3 VaP)8 = (3 aM)8 = 27 Jb^ = 27 Vo^S*.
Raise to the indicated power
3. {Vsy. 7. (v^)'. • 11. {V^^
4. {5</3y. 8. (^*. 12. (2V^6^*.
5. {</i2y. 9. (V^«. 13. (4^;/a"=^)*.
6. (V32)\ 10. (2^/a^*. 14. (xVyy.
15. Extract the cube root of SVo^.
v^8Vax8 = (8 aW)^ = 2 a^x* = 2^^.
§405.] RADICALS. 228
16. "v^VeJ. 19. V26\/6. 22. ^^^/^^.
17. "V^VSl. 20. "V^Vo^. 23. ^aWa.
18. Vl25V8. 21. Vo^Va?. 24. ^^o^.
Square Roots op Binomial Surds.
403. A binomial one or both of whose terms are surds is
called a binomial surd. Thus, a \ V6, Va — b, and Va ± V6
are binomial surds.
The finding of the square root of a binomial surd which is the sum ol
a rational number and a quadratic surd, as a i: Vb^ is of sufficient utility
to justify a brief treatment here.
404. Since ( V5 ± V sy = 5 ± 2 VlS + 3 = 8 ± 2 VlS, then,
conversely, Vs ±2Vi5=V6 ± V3; and hence 8 ± 2^/l5 is
a perfect square. It is thus seen that some binomial surds
of the form a ± Vb are perfect squares.
405. A formula for finding the square root of such binomial
surds may be obtained as follows :
(VahVS ± Va  Viy = 2a± 2Vc?^^.
.'. Va+V6+VaV6=V2a + 2V^^, (1)
and Va+V6VaV&=V2a2V^^. (2)
Adding (2) to (1), member to member, and dividing by 2, we
have
and subtracting (2) from (1), and dividing by 2, we have
(3)
V^3;^ Jg+V^'ft JgVa'ft.
2 2
(4)
224 ALGEBRA. [§ 406.
406. It is evident from Formulas (3) and (4) that the square
root of a ± Vft may be readily found if a? — h is a perfect
square. This may usually be determined by inspection, and
the square root of the binomial is then found by substituting
the values of a and Va^ — h in the proper formula. The
method is not practicable when a^ — b is not a perfect square.
1. Find the square root of 6 4 VlT by formula.
Va2  6 = V6211=V26 = 5,
whence V^Tv!T=V^ + a'^ = V + A = i>^ + i^
2. Find the square root of 17 — 4 Vi5.
^174Vl5 = Vl7V^; V172  240 = VS = 7 ;
whence Vi^ ^4^/l^=J^^^^^J^^^^:^ = Vl2^^/E==2y/S^^
Find the square root of
3. 64.V2O. 8. 7 + 2VlO. 13. 6a\2aV5.
4. 10+Vi9. 9. 11V72. 9
14.
5. 7 + 4V3. 10. t^+VJ. 6 + 2V6
6, IIV2I. 11. fV7. „ 121
ID.
7. 21 + 4V5. 12.  + iV24. 9Vi7
407. The square root of a binomial surd may also be found
by first so reducing the surd term that its coefficient is 2 ; then
separating the rational term into two parts such that twice the
product of their square roots equals the surd term ; and then
connecting such square roots with the sign of the surd term (§ 339).
Thus
(1) V6+vTi = VV +2Vv+^=V v: + V=^V22 + ^V2.
(2) Vl74Vl5 = Vl22V6()f5 = Vl2V5 = 2V35.
Find by this method the square root of 3 to 15 above.
§409.] RADICALS. 226
Another Method.
408. If a ± y/b = c± V5, and a and c are rational, and Vft
and V^ are surds, then a = c, and b = d.
For, if possible, let a = c  a? ;
then, substituting, c + x± y/b = c ± V5 ;
whence x ± V6 = ± V5.
Squaring, a^ ±2 xy/b h 5 = d ;
transposing, o^fft — (Z = :f2 oj V&,
which is impossible, since a rational number cannot equal a
surd; and hence
a = c, and V6=Vd, or 6 = d.
409. The square root of a binomial surd of the general form
a ± VS may be found as follows :
Suppose Va ± Vft = VS ± V^ ; (1)
squaring, a ± y/b = a? ± 2 V^ H y.
Hence, by § 408, we have
a? + y = a, and 2 V^ = Vft ;
or 3^ + y = a, and 4iBy= 5. (2)
The value of v a ± Vft can now be found by finding the values
of X and y m equations (2), and substituting the same in (1).
16. Find the square root of 15 + 2 V56.
Let Vl6 + 2y/m = Vx + Vy ;
squaring 16 + 2V56 = « 4 2Vicy  y ;
whence « + j/ = 16, and 2Vay = 2\/56;
or X 4 y = 16, and ay = 56. .•. a: = 7 ; y = 8.
Hence Vl6 + 2V'66 = V? + VS.
Find the square root of
17. 7h2VlO. 19. 11 + 2V30. 21. 16 + 2 VSK
18. 6+V20. 20. 126V3. 22. 62/^.
whitk's alo. — 15.
226 ALGEBRA. [§ 410.
EQUATIONS INVOLVING RADICALS.
410. An equation in which the unknown quantity occurs
under the radical sign is called a radical equation.
411. The more common methods of solving radical equa
tions are indicated below.
1. Solve the equation V«* — 15 f a? = 5.
Transposing jc, Vac* — 16 = 6 — ac ;
»juaring, a?  16 = 26 lOz + afi;
transposing, 10 x = 40 ;
whence * « = 4.
2. Solve the equation VaJ + 7 = 7 — a^.
Squaring, a; + 7=49 — 14>/5c + x;
transposing, 14 v^ = 42 ;
dividing by 14, Vx = 3 ;
squaring, x = 9.
3. Solve the equation V« — a= Va5*4«aj.
Raising Vx a to 4th degree, v^(x — a)^ = v^x* + ax ;
raising to 4th power, x* — 2 ox + a* = x* + ox ; "
transposing and changing sign, 3 ox = a* ;
whence x = — = .
3a 3
Solve
4. Vaj4 = 6. 11. V5^T4=V3a + 2.
6. V7a;38 = 6. 12. V4aj 19 = 2VS 1.
6. Var^16 = i»2. rs. V^T5+V« = 5.
7. Va?l=V» — 9. . y x — 1
^ ^ 14 Vo; + 3 = >  '
8. Vajh3 = V3ajl. Va?3
9. 5V«^^ = 3Va + 5. 15. Va; + 12 = 2+V5.
10. \^4a; + 3 = 3. 16. V4+Va^ = a
§411.] RADICALS. 227
17. VaJ + 3a = 3Va. 26. a + » = VaJ^ + Sooj — 2a*.
18. V^Ta^=6«. o^ g^ ^__ 1
^ f . — ~ ~~ JC —
19. V^Ta' = V» + a. Va^l Va^l
20. V« — V2=Va? — 2. gg Va;4«+ V^ _g^
21. V» + 4a6=V5 + 2a. VaJ + a V®
22. 2 V» — Va = 2 Va? — a. 29. Var* — a^x = ^/ai^ + 6V.
23. VS+^V^=^^=V26. 30^ VS + 2=Vi_V^T21.
24. Va? + a = Va + Va: — a. / « . o
4 31. •V^^ + 3 = ^±^.
25. v^+2+V^^=7== 1 VaJH2
Va;+2
Miscbllahbous exbrcisbs.
Simplify
1. f 2^ V^ — i aa? V»^. 11^ Va; — 4 — 5
 :^^^^^ (.+,)^ ^^^ (a. + .)V^.
3. VaiB»4a''ar+4a»+Vl6^. ^'g + y 
4. (^</)xa/. 13. Vll + 4 V6.
\2__V3 \6 jg V 9 V5 + 21.
6. Va_V6xVa+V6. ^^ V2I  4 V6.
7. ^±4. , 3
8.
'"+^ "" V2+V3+VB
wab — ax 2
hxVah • V2+V31'
Va; + 1+Va;1 1 1
9. , , 19. 1 ■—
VSTTVa;1 ' V5V2 V5+V2
10. V^^_W^. 20. — ? ^^ —
Va? — y + Va + y VT + I V7 — 1
228 ALGEBRA. [§ 412.
21. V6x</i2x</Sx</Ux^/lS0.
22. (x{V^+Vf)(^V^Vf).
23. (a^\'Va + ^/b){a^—^/a■^/b).
24. Va^ — &^ H(Va — ft X Va — 6)
Knd the value of a; in
25. ^L—+ ^ =^. 27. V^^T^+V^^ ^,.
Vl— a;+l Vl+a;— 1 ^ ■y/a^x — 'Va — x
26
v^^
29. ■\/x—a—y/x—b=Vb — Va, 30. 2 V5— V4 a?— 11=1.
IMAGINARY NUMBERS.
412. Since (± a)^ = + a^, (± a)* = + a^ and (± a)^= + a«"
(§ 316), it follows that every even power of any number is
positive; and hence the even root of a negative number is im
possible, and hence such a root can only be indicoUed by the
proper sign.
413. The indicated even root of a negative number is called
an imaginary number. Thus, V—a and V— a^ are imaginary
numbers.
In distinction from imaginary numbers, all other numbers
are called real numbers.
An imaginary number is more properly called a nonreal number,
numbers being thus classified as real and nonreal ; but the term imagi
nary number is in almost universal use.
414. The indicated square root of a negative number is
called a quadratic imaginary number.
It may be shown by the methods of higher algebra that every imagi
nary number may be expressed in the terms of a quadratic imaginary
number, and hence we shall consider herein only quadratic imaginary
numbers.
§417.] RADICALS. 229
415. Since
V^r^* = Va*x(l)=V^xV^^=aV^
and V— a =Va x(— l)=Va x V— l = a^V— 1,
it appears that any imaginary square root may be changed
to the form aV— 1 or a'V— 1, in which a and a* are reoi
numbers.
The radical V— 1 is called the imaginary unit.
Reduce to the form aV— 1
1. V4.
2. V^=^. ^ 16
• V
9sc«
8. V^^256. rr^
^•>/?
«• \^ 10. V(a + &)*.
416. An indicated square root of a number is squared by
simply removing the radical sign, as ( Va)^ = a, and (V— a)^
= — a; and hence the square of the imaginary unit V— 1
is — 1.
417. The successive powers of the imaginary unit are as
follows :
(V^::i)8=(V3i)2xV^=T = ixV=l = V^^;
(V^*=(V^^)»x(V^^ = lxl = + l;
and so on.
230 ALGEBRA. [§ 4ia
Hence, if n denotes any positive integer,
(V^)** =[(V=^)*]"=(1 X l) = + l;
.. ( v=^)^+' = + 1 v^^ = v^=i: ;
.. (V^^)^+* = + lxl = l;
418. It is thus seen that the first four powers of V— 1 are
respectively V— 1, —1, — V— 1, and +1; and that these
numbers occur in the same order for the powers of V— 1
whose exponents are 4w + l to 4w + 4 inclusive.
Addition and Subtraction.
419. Imaginary numbers may be added or subtracted in the
same manner as other radicals (§ 389).
1. Add V^^ and V^^25.
2\/^=T + byT^ = 7 V^.
2. Simplify V 144  V^TIq _ V^Zie.
V^^U4 = 12\/^T
Simplify
4. 3V162V25. IT ifir T
. ^r—T ./ 9 9 V^^V^=4^+V^=^9^.
7. V=^+V:^36V=65, yi a' M a' M a«'
§ 420.] RADICALS. 231
Multiplication and Division.
420. The only special difficulty in multiplying or dividing
imaginary numbers is in the sign of the product or quotient.
Thus, V^^ X aA^ is not V2x(~3) = V6, as in the
multiplication of radicals (§ 393), but V6 X (— 1)= — V6.
This difficulty is avoided by first reducing the imaginary
numbers to the form a V— 1, as below.
1. Multiply V^ by V^.
V^Te = Vo X V^^
>/^ X V^ = VSO X ( V^=T:)2 = V30 X (  1) =  V30
2. Divide V^^^^ by V^^Ts.
V I2;V^ = V4 X l=\/4
The same result would be reached in Example 2 by proceedinf;: as in the
division of radicals (§ 397); thus, V372 r V3"3= V 12~ ( 3) = Vi.
A reduction of terms to the form aV— 1 is necessary in division only
when one term is imaginary, and in multiplication when both terms are
imaginary.
Multiply
3. y/ZIs by V^r5. 7. V^ by V^=^.
4. 5V^ by 3V^^. 8.  3 V^=^ by 5 V^^.
5. 12V^^ by 6V^. 9. V^^ by V^^ by V^.
6. 8V16 by 16V^^. 10. V^ by V^ by V^.
Divide
11. V=^ by V3. 14. 4.V^^ by  SV^^.
12. V^^ by V— 5. 15. Va6 by V— a.
13. 6V3 by 3V^^. 16. 3V^=^ by 2V4a.
282 ALGEBRA. [§ 421.
421. The form a ± 6V— 1, in which a and 6 are real, is
called the typical form of imaginary numbers.
422. An imaginary number in the form a ± 6 VT is called
complex, and one in the form aV— 1 or V— a is called a pure
imaginary.
423. Two complex numbers which differ only in the sign of
their imaginary term are said to be conjugate (§ 395). Thus,
3— V— 2 and 3+V— 2 are conjugate, also x + y^/—l and
« — yV— 1.
424. Both the sum and the product of two conjugate com
plex numbers are real. Thus, a + 6V— 1 {a — 6 V— 1 = 2 a,
and (a + 6 V^(a  b^/^^)= a^ ( b^= a^ + ^.
It follows that the sum of two squares can be factored by introducing
the imaginary unit. Thus, a^ \ b^ =(a{ bV— l)(a — bV— 1).
426. An imaginary number cannot equal a real number;
for, if a? = V— a, ic^ = — a, which is impossible, since a positive
number cannot equal a negative number.
426. If two complex imaginary numbers are equal, their
real parts and their imaginary parts are respectively equal ;
for, if a\bV —l=X'\y^/ —1, transposing, a— aj=(y— 6)V— 1,
which is impossible unless a = «, and y = b. For, if a > or <
Xf then we have a real number equal to an imaginary number
if y > or < 6, and to it y = b, both of which are inq)ossible ;
but, if a = Xj then y = b, otherwise we have equal to an imagi
nary number, which is impossible. Hence a = x, and y = b.
427. If the denominator of a fraction is a complex imaginary
number, the fraction may be rationalized by multiplying both
terms by the conjugate of its denominator, as in § 400. Thus,
l( 3fV2) _ 3fV2 _ 3+V2
3V^::2 (3V2)(3+V^2) 9(2) 11
§ 427.] RADICALS. 288
Simplify
17. (3 + 2 V^=^)(3  2 V^=^).
18. ( V32  3 V^^XV^^ 4 3 V^).
19. a2 + 6V^^+(a + V^^)(aV^^)
20. (aV^^ — cV^^)(aV^+cV^^).
21. (V:r3+v"i:2)(V^2V^.
22. (i 4 1 v^)a  1 v^).
23. f.^Ya:4.^
24
25
26
. (V^r3^_v^r2)^(V^r3Vir2).
27. ^ ^  28. — ^
4  3 V 2 a  aji,/^=T
^^ 3V32 42V£3 , 3^ V^ + V^ .
31. 3 2
52V^^ 5 + 2V^=^
o„ a 4 aV— 1 a — O/'V— 1
a — xV— 1 a 4 aV— 1
a— V— 1 a;4V— 1
34. 2 v^=n: 4 2 V^^  3 v^^.
35. V^^+V^^12 4(V^(V^2Vi:8).
Expand
36. (3V^l 39. (a4V^/(aV^)2.
37. (V^^2V^^)l 40. (xV^^ { yV^y.
38. (5V=^V^^)'. 41. (l + VZi:)^^(l_V^)*.
234 ALGEJBIiA. [§ 42&
CHAPTER XIV.
FRACTIONAL AND NEGATIVE EXPONENTS.
428. The definition of exponent already given (§ 29) applies
only to exponents which are positive and integral; and the
laws of exponents established have also been considered as
referring to exponents which are positive and integral.
429. It remains to explain the meaning of fractional and
2
negative exponents, as a^ and a~^, and also to show how oper
ations are performed with numbers which contain such ex
ponents.
FRACTIONAL EXPONENTS.
430. It is assumed that the fundamental law of exponents,
a** X a" = a*'*"'* (§ 84), is true, whatever may be the value of m
and 71.
Hence, if a"* x a*" = a"+**, then a^ x a^ = a^^^ = a ; and
since a'^ x a^ = a, a^ = y/a ; that is, a^ denotes the square
root of a.
431. Since a^ x a^ x a* = a*"*"*"*"^ = a, a^ = J^; that is, a*
denotes the cube root of a.
Ill 1 1
Since a" x a" x a** ••. to n factors =:a, a*" = y/a] that is, a*
denotes the nth root of cp.
§435.] FRACTIONAL AND NEGATIVE EXPONENTS. 236
m m IK iM fM
~" ^ ■"• ~~Xfi ~"
Since a" x a* x a* ••• to n factors =0** = a* a" = \^a"*;
and hence a" denotes the nth root of the mth power of a. Hence
432. The numerator of a fractioncU exponent denotes the power
of a number, and its denominator denotes the root of that power.
Thus, x^ denotes the fourth root of the cube of x,
111 1 m M
Since a*» x a* x a* to m factors = a* ^a"*, a"^ = (x/a)** ; and hence
a» also denotes the mth power of the nth root of a.
But this is strictly true only when the root is arithmetical or positive.
For example, Vo* = ± a^ ; but (Va)* has only one value, + a*.
433. Whatever the value of a may be, a° = 1 ; for a* ^ a*
= a"" = a°, and a'^ia'* = 1. .*. a® = 1 (§ 123).
NEGATIVE EXPONENTS.
434. If the formula a"* j a" = a""^ is true for all values of
maiidn,then a' ^ a' = a'' = a'^
But a* 5 a* = — = — •
a* a*
.. a~^ = — ; that is, a~^ denotes the reciprocal of a\
Likewise a" s a^ = a*"~^"* = a""* ;
but a* ^ a** = 7 = —
a^"* a"*
.*. a~** = — ; that is, a"* denotes the reciprocal of a*,
a**
435. It is thus seen that any factor may be transferred from
the numerator of a fraction to its denominator , or from the
denominator to the numerator, provided the sign of its exponent
be changed.
Thus a^b^' ^aZ^2JL. «'^' ^"V .
286 ALGEBRA. [§435.
Express as integers
1 — :; — ::• 4. — : • 7.
a'%'^ ajiy^f
6. ""' . 8. ^y^
2. ^^ 6. a&^ ^ __
X *^ *2
3. — _ — . 6. ^ 9. '
Express with positive exponents
8«2
10. «*' .
12. *■v^
14.
11 «"'^"'
13. ^^.
ary*
15.
Express w
ith fractional exponents
19. ^16 aV.
22.
16. v/^.
^a'hh'i.
17. y/a^T^y,
20. ^X
ft 1 — ;
>
23.
^a«6» + VaV.
18. ^^27.
21. V^^
■>/d\
24.
^/'a^ft^ : a^6*.
Express with radical signs and positive exponents
25. a'^ft^.
27. SM.
29.
31. — i
26. a"Vl
28. 2cr^V.
30.
3_2
4»
32. <^h%\
Find the value of
33. 8*.
36. 4"*.
39.
64i
42. 36*x3».
34. 16*.
37. 16^
40.
lOOH
.
43. 25* 4 27*
36. 27^
38. 9^ X 27^.
41.
1000*.
44. 9'*X36*.
§438.] FRACTIONAL AND NEGATIVE EXPONENTS. 237
OPERATIONS WITH FRACTIONAL AND NEGATIVE
EXPONENTS.
436. The several laws of exponents may be expressed by the
following formulas :
(1) a"* X a'' = a"*"*"*. (4) (a 5 ft)* = a"* t 6*
(2) a s a" = a*". (5) (a"')" = a"'".
(3) (aft)" = a'ft*. (6) Var' = a'^.
437. These formulas being true for all values of m and n
(§ 430), it follows that the rules previously given for opera
tions with numbers whose exponents are positive integers, also
apply to operations with numbers which contain fractional or
negative exponents.
438. Operations with radicals may be readily performed by
substituting fractional exponents for the radical signs, and
then proceeding as with rational numbers (§ 402). By such
substitution the operations with radical numbers may often be
much simplified.
1. Multiply 5 a6^aj% by Za^b^a^y^.
The product is found by multiplying the coefficients, and adding the ex
ponents of the like literal factors (§ 109). The product is 15 cfib^x^y^.
2. Divide ofix^y^ by 3 a'aj^t/*.
3. Multiply a + a*6* + & by a*  6*
a {■ ah^ {■ h
 ab^  ah  b^
a* 6*
238 ALGEBRA. [§ 438.
4. Divide
a?* — 3 x^y~^ \ 3 x^y^ — y"^ by x^ — 2 x^y"^ + y~^.
x^  3xiy i + 3xV*  y~^
x^ — 2 x*y"^ + y ^
aji _ 2 x^y i + x^yi
x*^ — y '
 x*y"^ + 2 x*y"^  y~i
 Q^y~^ + 2 xiy"i  y"i
Multiply
6. a^ + 63 by a^ — 6*. 9. «» + a;%^ 4 y^ by a;^ — y^.
6. a* + 6""* by a* — 6"*. 10. a;"* + »"* + ! by. a;"* — 1.
7. aj + «"^ by a; 4 «"^ H. a + a;* + 2 by a; + a;* — 2.
8. x^\y'^ by x~^^y^. 12. aj*+a*+l by a?*— ar*+l.
Divide
13. ah by a* + 6*. 16. 27 x^ ^^y'^ by 3aj"* f 2y\
14. a H 6 by a^ + 6^. 16. a;^ — y^ by a;^ + 2^».
17. a* — a6* + «*&  &* by a* — 6^
18. a; + a?^2/^ + y by a?^ — a;^y* + y^
Extract the square root of
19. 4a — 4aMH&. 21. a?^ — 4a;* 42a; + 4a;^ + a;*.
20. aj^+2a;"'V*+y"* 22. a;'*— a;V^4 ^^"^" — a?"V"^4y^.
4
Simplify 23. (a3)2x(a^^. 26. A/(aW^^^W^.
24. (3 2 + 1)* 27 («V) X (pj:)^.
1 1 (wl)
§442.] QUADRATIC EQUATIONS. 289
CHAPTER XV.
QUADRATIC EQUATIONS.
439. An equation which, when cleared of fractions, contains
the square of the unknown number, but no higher power,
is an equation of the second degree. An equation of the second
degree is called a quadratic equation. Thus, 4a^ = 16, and
3 a^ — 4 a? = 15, are quadratic equations.
440. A quadratic equation may be reduced to the general
form aa? {hx{c = 0, in which a, 6, and c are known numbers.
The known numbers, a, 6, are called the coefficients of the
equation ; and the term c, the constant term.
The term c may also be made a coefficient by writing the equation in
the form of ax^ + &x + cico = 0.
441. When the coefficient h is zero, the equation becomes
as? + c = 0, and is said to be incomplete, because the first power
of X is wanting. When no one of the coefficients is zero, the
equation is said to be complete. Hence
An incomplete quadratic equation contains only the square of
the unknown number.
A complete quadratic equation contains both the square and
the first power of the unknown number.
An incomplete quadratic equation is also called a pure quadratic ; and
a complete quadratic equation, an affected quadratic.
442. The root of an equation is the value of the unknown
number in it (§ 138). When a root is substituted for the
unknown number, it satisfies the equation; that is, reduces it to
an identity.
240 ALGEBRA. [§ 443.
INCOMPLETE QUADRATICS.
443. An incomplete quadratic equation is reduced, if neces
sary, to the general form as? = c by the same transformatioDs
as those employed in the solution of simple equations.
444. An equation of the form aa? = c is solved by dividing
both terms by a, and then extracting the square root of both
members.
Thus,
aa? = c;
dividing by a,
a
extracting square root,
x = ±
445. The substitution of +\l or — %/ for x will satisfy
the equation (§ 442) ; and hence an inwmplete qxiadratic equation
has two rootSf numerically equal, but having opposite signs.
446. A root which can be exactly found, as V9, is a rational
root.
A root which can only be found approximately, as V5, is an
irratiortal root, or surd (§ 371).
A root which cannot be found either exactly or approxi
mately, as V— 9, is an imaginary root (§ 413).
1 . Solve the equation ^±? + ^^ = ??.
^ aj3 aj + 3 10
Clearing of fractions, 10 (x + 3)2 + 10(x Sy = 29(x^  9) ;
expanding, 10x2 + 00 a; + 90 + lOx^  60a; + 90 = 29x2  261 ;
transposing and simplifying, x2 = 49 ;
extracting square root, x = ±7.
§446.] QUADRATIC EQUATIONS. 241
2. Solve the equation ^^^=2aj2 7.
Clearing of fractions, 6^2 _ 5 _ 4 3.2 _ 14 .
transposing, etc., x^ = — 9;
whence x = ±^/^^ = ±SV^ri.
Since the square root of a negative number is imaginary (§413), the
value of X can only be indicated.
3. Solve the equation x — ^^ — 3 =
V?^3
Clearing of fractions, xy/x^ — 3  (a;2 _ 3) _ 2 ;
transposing and uniting, xy/x^ — 3 — x^ — 1 ;
squaring each member, a^ — 3a;2 = a;4 _ 2 x^ + 1 ;
transposing and simplifying, x^ = — \\
whence x=± V^^.
Solve the equations
4. 7a? = mi. ^^ a;42 x2 ^^
6. 11 ic* 9 = 35. ' x~2 x + 2
4
X
7. i^5 = 7. ^^
16. ^+ 1 1
4 a; — 3
10. ^=10. 2aj2^ 4a2
11. x\=2x. 19 —
2a; x\2
3aj aj45 o
13 ^4 ^ ar^4ll 01 a?45 a?5 _ 15
3 4* • aj5"^aj + 5"" 4*
white's alo. — 16
242 ALGEBRA. [§ 446.
22. VT+^x l 25. ?1^2L=a^,
a* 4 ax
23. ^2 = — !?_. 26. ^±60^^_^jzl2.
or — n jf — m 6 3
^*« Vaia V 27. — ' = —
■ya — x x — a x + a c
a? 4 6 a? — 6
29. ^ ^'^ 1
30.
31. Two numbers are to each other as 3 to 5, and the differ
ence of their squares is 256. What are the numbers ?
Suggestion. Let 3 x and 6 a; be the numbers.
32. Two numbers are to each other as 2 to 7, and their
product is 126. What are the numbers?
33. A rectangular field contains 6 acres, and its sides are
to each other as 3 to 4. What is the length of each side ?
34. The length of a rectangular field is 2^ times the width,
and the field contains 9 acres. What is the length of each
side?
35. The sum of two numbers is 16, and their product is 60.
What are the numbers ?
Suggestion. Let 8 + x and 8 — a; be the numbers.
36. A father's age is to his son's age as 5 to 2, and the
product of their ages is 640. What are their ages ?
37. Two numbers are to each other as a to &, and their
product is c^. What are the numbers?
38. A son's age is to his father's age as a to 6, and the
product of their ages is m. What are their ages?
§449.] QUADRATIC EQUATIONS. 248
COMPLETE QUADRATICS.
447. There are, as will be seen, several methods of solving
complete quadratic equations; and it is advisable for pupils,
especially those who expect to enter higher institutions, to
acquire a knowledge of those in more common use, though
in practice it may not be best to use more than two of them.
Method op Solution by Factoring.
448. A product is zero if any one of its factors is zero
(§ 22) ; and, in order that any product may he zero, at least one
of Us factors must be zero.
For example, if (x — 2) (a; 4 3) = 0, then a5 — 2 = 0, or
a? 4 3 = 0. But if a;  2 = 0, a; = 2; and if a? f 3 = 0, a =  3;
and, since either value of x thus found will satisfy the equation
(x — 2)(x f 3)= 0, its roots are 2 and — 3.
449. It is thus seen that any quadratic equation whose first
member is a product, and whose second member is zero, may
be solved by equating each of its two factors to zero, and solving
the resulting equations.
It is here assumed that the factors are finite.
1. What are the roots of the equation (x + 3)(a: — 1) = ?
Equating the factors to zero, a: + 3 = 0, and x — 1 = ;
whence x = — 3, and x = 1.
Hence the roots are — 3 and 1.
Find the roots of
2. (a;7)(aj2)=0. 6. (« + i)(aJ i)=0.
3. (aj + l)(aj5)=0. 7. (a;  6)(a: h c) = 0.
4. (a?6)(a;4)=0. 8. (a  a)(aj V6)=0.
5. (a?  i)(aj + 2)= 0. 9. (aj+V^)(ajVa)=0.
244 ALGEBRA. [§ 450.
450. A quadratic equation of the form a? \ bx \ c = may
be readily factored by § 202, and its roots thus found, pro
vided c is the product of two rational factors whose algebraic
simi is b.
For example, let aj* + 3a; — 10 = 0.
Factoring by § 202, (x + 5)(x  2) = ;
equating factors to zero, a? + 5 = 0, and « — 2 = ;
whence x = — 5i and x = 2.
Hence the roots of the equation are — 6 and 2.
10. Solve the equation a^ — 7x = — 12.
Transposing  12, a;^  7x + 12 =
factoring, (a; — 4) (x — 3) =^
equating to zero, oj — 4 = 0, and x — 3 =
whence x = 4, and x = 3.
Hence the roots are 4 and 3.
451. A quadratic equation of the form oo* + fta? + c = may
be factored by § 205 or by the general formula, § 208, and its
roots be thus found.
11 . Solve the equation 3 a* + 7 « = 6.
Transposing 6, 3x2 + 7x  6 =
factoring by § 206, (3 x  2) (x + 3) =
equating to zero, 3x — 2 = 0, and x + 3 =
whence « = f » and x = — 3.
Hence the roots are f and — 3.
Solve by factoring the equations
12. a^aj= 12. 18. a^ 10a; = 56.
13. a^4:x = 4:5. 19. o^ f 15 a? =  26.
14. a^10x=z21. 20. ar»  15 a? = 154.
15. ar*  12 aj =  32. 21. 2a*+6a? = 20.
16. a:* + 3aj = 28. 22. 3a* 7a? = 40.
17. ar^ + a = 56. 23. 5a*+27a?=18.
§454.] QUADRATIC EQUATIONS. 245
24. 6a?7aj = 20. 28. a? + 2a^ ^Sax = 0.
25. 4ar*— lla: = — 7. o« «2 / b\ , ,
26. 4ar^23a; = 15.
27. aj^ + aaj + 6a? = — a6. 30. aj^+(Va4V6)aJ= — Va5.
452. ^ny complete quadratic equation may be solved by
factoring, but the method has no special advantage when the
factors cannot be determined by inspection. When the factors
can be thus readily determined, no other method of solution
need be used.
453. When a quadratic equation can be readily factored by
inspection, its roots may be written at once without equating
the factors to zero.
Thus, the roots of aj^ — 2a; — 24 = are seen to be — 4 and 6.
Write, without equating factors, the roots of
31. ar* + 6 a; h 9 = 0. 34. ar^ 7 a; 60 = 0.
32. ar^ 5 a; 46 = 0. 35. 2ar^ + 10a; + 8 = 0.
33. a;2_^3a._10 = 0. 36. 3 a;^ 12 a; 15 = 0.
454. Conversely, if the two roots of a complete quadratic
are given, the equation can be formed by findiiig the product of
the two binxymial factors whose first terms are x, and whose second
terms are respectively thei;wo given roots with signs changed, and
then writing the product thus found equal to zero.
Thus, if —2 and 3 are the two roots of a complete quadratic,
the equation is (x f 2)(x — 3) = 0, or a;* — a; — 6 = 0.
It can be shown that a;^ — a; — 6 = is the only equation whose roots
are — 2 and 3 with no other roots, but the proof is too difficult for
insertion here. 3a;2_3a; — 18 = is reducible to x^ — x — Q^Q^ and
hence is no exception.
37. Form the quadratic equation whose roots are 3 and —5.
a; = 3, and » = — 6.
Transposing 3 and — 5, oc — 3 = 0, or« + 5 = 0;
whence (x — 3) (a; + 5) = ;
/. «3 + 2 a;  16 = 0.
246 ALGEBRA. [§ 455.
38. Form the quadratic whose roots are f and —J.
as = j, and x = — J.
Transposing, oc — f = 0, and as + J =
whence (« — i) (« + i) = <^
... a;2ia;i=0
clearing of fractions, 6 x^ — a; — 2 = 0.
455. The required equation may be written at once by
changing the signs of the given roots, and then making their
algebraic sum the coefficient of aj, and their product the third
term.
Thus, if the two given roots are a and , the equation is
b
«^+(«+?>f=«
Form the quadratic equation whose roots are
39. 1 and — 5. 44. — 4 and 0.
40. — 3 and — 4. 45. —VE and VS.
41. — 2 and ^. 46. V3 and — V3.
42. I and— J. 47. 2 — V3 and 2 + VS.
43. — i and — 4. 48.  and — a.
8 2 . ^
Solution by Completing the Square.
456. Any complete quadratic equation may be reduced to
the form of ±hx = c by dividing, if necessary, both members
by the coefficient of a^,
457 If the square of ^ of 6, or (  ) , be added to both mem
bers of a^ + bx = c (1), called completing the square, the equa
tion becomes
The first member of (2) is now a perfect square, the ex
tremes being each a perfect square, and the middle term twice
the product of their square roots (§ 186).
§457.] QUADRATIC EQUATIONS. 247
Extracting the square root of each member of (2), we have
"■2
and transposing + > a? = —  ± v/c + —
It is thus seen that the equation has two roots,
 1 +>K»^ !>/<
It follows from § 454 that the first member of the equation x^\bx — e
= is the product of x + ^ }Jc + —\ and ac + / Jc + — V
1. Solve tte quadratic 3 aj* — 12 a; = 15.
Dividing by 3, x^ — ^x = 6;
completing the square, a;^ — 4x + (2) 2 = 6 + 4 = 9;
extracting the square root, aj — 2 = ± 3 ;
whence x = 2±3 = 5 or — 1.
2. Solve the quadratic a* 4 11 a; = 33.
Completing the square, x^ + n a; _^ (i^)2 = iji + iji = aja.
extracting square root, « + ^ = ± r^ ;
whence x =  J^ ± ^ = 2} or  13J.
3. Solve the quadratic ^^Lzl _ i^±2 = 2 aj  3.
X 5
Clearing of fractions, etc., — 14 x^ + 28 x = 35 ;
dividing by  14, x^  2x =  f ;
completing the square, x32x + l=f + l=};
extracting the square root, x — 1 = ± V— f ;
... x = l±v^ = l±i\Ar6.
The two roots are imaginary, as they will always be when c + — is
negative (§ 413). ^
248 ALGEBRA. £§ 458.
Solve by completing the square
4. a^8a; = 15. 7. iB*12a; = 45.
6. aj» + 12a; = 20. 8. jb*  20 a? + 19 = 0.
6. «»f 4a: = 21. 9. a^ + 18 a?  88 = 0.
10. 2a^10a? + 6 = lla?»*30.
11. 3x^llx = 20.
12. 5a:»aj + 19 = 3x» + 15«15.
13. 7a^12aj = 580.
14. (3a;410)(3a;47)=2(22a;a^.
15. 2ic2 + 94a;4420 = 0.
16. 3aj2  52a? 4118 =(5 2a;)(3a; + 2).
17. 2(a;3)(a:4)=aj*~25.
^ ajl32 39 ,^ 5a; + 3^7a;f2
ar^2a;20 2a;4l 8a;4 lla?3
20. ^+7=?. 21. T^ + ^^^ = i^.
2ar X 4 4 2
22. 2(a:  2)(a:  3) = (a? 4)(aj 3) +10.
23. 35\(a^\50)=a^^i(a^10).
458. To solve a complete quadratic by completing the
square,
Reduce the equation to the form a^ ^bx=c.
Add the sqtiare of one half the coefficient of xto both membersy
thus completing the square of the first member.
Extract the square root of each member of the resulting eqiLOiion,
and then find the two values of x.
The problem of solving a complete quadratic equation is thus reduced
to that of solving two simple equations.
§459.] QUADRATIC EQUATIONS. 249
Other Methods op Completing the Square.
Note. The four sections following (§§ 469462) may be omitted by
beginners.
459. Instead of dividing both members of the equation by
the coefficient of aj^, it is sometimes more convenient to multiply
both members, if necessary, by siich a number as mil make the
coefficient of x^ a perfect square, and then complete the square
by adding to ea^h member the square of the quotient obtained by
dividing the coefficient of x by turice the square root of the coeffir
dent of x^.
Take, for example, the equation 3 oj^ — 10 a; = — 3.
Multiplying both members by 3, we have,
9iB230a? = 9;
(30 \^
o X ^J
9aj2_30a;452 = 9 + 25 = 16;
extracting the square root, 3 a; — 5 = ± 4 ;
transposing and uniting, 3a5 = 5±4 = 9 or 1;
.. a? = 3 or \,
When the coefficient of a? is already a perfect square, the
square may be at once completed as above.
1. Solve the equation 18 a* — 15 a; = 42.
Multiplying both members by 2, 36 x^ _ 30 a; = 84 ;
completing the square by adding (f ?)^ =(l)^»
36x2 _ 30X +(1)2 = 84 +(f)2= 161 ;
extracting the square root, 6 x — f = db V" ;
transposing and uniting, 6 x = f ± y = ^^ or — 7 ;
.'. x = 2 or — 1.
Solve as above the equations
2. 8aj2~12aj = 80. 4. 10 ar^ + 2 =  12aj.
3. 3aj28aj + 4 = 0. 6. Ta^ _ 28a; =  21.
260 ALGEBRA. [§ 460.
6. 9iB" + 9a5 + 2 = 0. 10. S2 a^  60 x = 272.
7. 27fiB*30ir = 48. 11. 25 aj = 6 aj* f 21.
8. 5a*~44aj = 9. 12. 60 ic*  27 = 15 a?.
9. 12 aj* = 24 a? + 420. 13. 18a^ 27aj 26 = 0.
460. The foregoing methods of completing the square involve
fractions when twice the square root of the coefficient of x^ is
not a factor of the coefficient of x ; and these fractions some
times necessitate much work, as will be seen by solving by
either method the equation 29 a^ — 31 a; = 54.
461 . The first member of any complete quadratic equation
may be made a complete square without fractions.
For, take an equation of the general form
tta^\bx = c, (1)
Multiplying both members by 4 a (4 times the coefficient of
a^), we have
4aV + 4a6a; = 4ac. (2)
Since 4 a6 5 2 V4 a* = b, the square of the first member of
(2) is completed by adding b^ (the square of the coefficient
of X in (1), the given equation) to both members ; thus,
4 a^a^ + 4 a^>a; + &^ = 4 oc + 6». (3)
Extracting the square root, we have
2aaj + 6 = ±V4acTW. (4)
Solving the simple equation (4),
— 6 ± V4 ac + &*
X = •
2a
462. Hence any complete quadratic equation may be solved
without fractions as follows :
Multiply each member by 4 tim£s the coefficient of x*.
Add to each member of the resulting equation the square of the
ooeffi^dent of x in the given equation.
§463.] QUADRATIC EQUATIONS. 251
Extract the square root of both members of this equation^ and
solve the resulting simple equation.
This method was first used by a Hindoo mathematiciaii, and for this
reason it is known as the Hindoo method,
14. Solve the equation 13 a/^ — 15 a? = 22.
Multiplying by 4.13, 4.13'ia;2 _ 4.13.16a; = 4.1322 = 1144 ;
completing the square by adding 15^,
4.132x2 _ 4.13.16X + 162 = 1144 _(. 152 = 1369;
extracting square root, 26x — 16 = ± 37;
transposing, etc., 26 x = 62 or — 22 ;
.*. X = 2 or — \l.
When the coefficient of x in the giyen equation is an even number,
the square of the first member may be completed without fractions by
multiplying both members by the coefficient of x2, and then adding to each
member the square of half the coefficient ofx in the given equation.
Solve the quadratic equations
15. 3a^42a; = 2. 20. 15x^Sx + l = 0.
16. I3a^24aj = 205. 21. 20aj« 54a?= 104.
17. 15a^207 = 24a?. 22. 11 a^  24 = 10 a?.
18. 10a^f23a; = 12. 23. 3 aj» + 2 a? = 56.
19. 21a* + l = 10aj. 24. 5a^S = 6x.
Method op Solution by Formula.
463. The solution of a complete quadratic equation of the
general form aa?\bx=c gives as the two roots
— b±V¥~+Tac
x = •
2a
Instead of working out the solution of every equation from
the beginning, completing the square, etc., we may write out
the roots at once by substituting for a, b, and c in the above
general formula their values in the particular equalion, as
shown below.
252 ALGEBRA. [§ 464.
1. Solve Uie equation as* — 11 a; = — 24.
In this case a = 1, 6 = — 11, and c = — 24.
Sabstitating these values of a, 6, and r, in the formnla, we have
_ 11±VTP"H4(24) _ 11 j:Vi2n^^96 _ ll ±V25 .
*~ 2 2 2 '
2
In snbstitnting, special attention most be given to the sigjis of the co
efficients.
2. Solve the equation 17 ic* + 8 a; = 21.
a = 17, 6 = 8, and c = 21.
Substituting these values in the formula,
^ 8±V64+1428 8±Vl492
34 34
^ ^  8 ^38.6264+ ^ ^^ ^^ 1.3713.
34
Since V1492 is an irrational number, the two roots of x are surds.
464. The correctness of the roots obtained may be verified
by substituting them for x in the given equation; but it is
usually better to verify by the aid of the principle stated
in § 466, that the sum of the two roots in the general for
mula = ^^^, and their product =
a a
Thus, in Example 2, above, 84Vn92 8Vi492^8
' ^ ' ' 34 34 17'
and 84Vn92^8.^Vi492^21
34 34 17
In using this method, a glance at the coefficients of the equation will
usually show the correctness of the results.
Solve by formula and verify the roots of
3. aj2  24 0? = 481. 6. ar^  24 a? =  119.
4. ar^ 41a? 348 = 0. 6. 3ar^lla? = 4.
§466.] QUADRATIC EQUATIONS. 263
7. 2ar2 + 38aj = 364. 11. 3a? llx20 = 0.
8. 4aj217a; = 42. 12. 2x'^4:X5 = 0.
9. 5x2 + 21 a; = 62. ^3 rnoi? { nx — 4: = 0.
10. 7x2^34^.^24 = 0. 14. mV + wxl = 0.
466. Assuming that the coefficients in the general equation
ox* \ bx = c are finite, and not zero, it can be shown, by
methods that properly belong to higher algebra, that it has
two roots, and only two.
466. The roots of a quadratic equation have the following
general properties:
I. Adding the two roots as expressed in the formula
^ — =— — — — 2^, it is found that their sum is ^^^^; and, multi
2a a ' '
plying them, it is found that their product is ,
II. If 6* + 4 ac be positive, y/b^ f 4 oc is real, and the
roots of the equation are real and unequal.
III. If 6* 4 4 ac be a perfect square, the two roots are
ra>tional ; if 6* f 4 ac be not a perfect square, the two roots are
irraJtioiud.
IV. If 6^ + 4 ac is negative, both roots are imo/ginary.
V. If 52 + 4 ac = 0, both roots reduce to •^— , and are thus
2a
equal; and it is then said that the equation has two equal roots,
LITERAL QUADRATICS.
1. Solve the equation (a — x) (a + x) = a (x — a).
Multiplying, transposing, etc., x^ + ax — 2 a^ =
factoring, (x — a)(x + 2 o) =
equating factors, x — a = 0, and x + 2 a =
. *. X = a or — 2 a.
ALGEBRA. [§ 40$.
2. Solye tiie equation 2* + m' ~ nx = »ii.
Complelmg the aqoaie, a*+ (»—■)*+ (^'^i^ V=**+ *^^^*'*
4 •
cxtiaeliDg the square root, ^^^ m * =^.J*±J!;
2 2
x = ?ill«±"*^
2 2
.. x = n or — m.
3. Solye by fommla the equation €12?— (a* — 5)x = a&.
2a 2a
.. x = =a,o>r = •
2a 2a a
Solve the following equations by any one of the above
methods:
4. a? — OiX — x = — a. ab
. ^ 11. 2 + ^ = a + 6.
^ff 12. x4a6 = (^ + ^X^^)
7. aaj«2aaj + 2 = 2. 13. ^ 5L_ = _«5_.
a + a5 a — x a — x
8 £±£ + — 2L_1 =
a a — x ' 14. a* («* + «*) = 2 rfoj+l.
9. a?(n + l)x = n. ^g aj»2aaj + te2a& = 0.
in «^ g'
^"' ^TT~a + l' ^®' ^— (« — & — c)a = ac + a6.
17. aj»(2c + 2(f)aj = 4c(i3(?(P.
4a&
18. (a + 6)aj* — 2(a — 6)aj =
19.
a + b
x\l m f 1
Va Vm
§467.] QUADRATIC EQUATIONS. » 266
HISCBLLANBOUS EZBRCISBS.
467. Solve the following quadratic equations, each by the
method that may seem best adapted to its solution. Verify
the results by seeing whether the sum of the two roots is ^^^^,
— c ^
and their product (§ 464); or by direct substitution.
a
19. H =a46.
2. a^aj182 = 0. »^ ««
3. 2««3aj6 = 0. 20. ? + ^ = l
« 9 o M a X a b
4. 3a^8aj = 4.
6. 9aj*12a; + 4 = 0.
7. 12aj226a; = 13.
8. .i^h2a^ = 3aa;.
21. a^ + a« — 6aj — a6 = 0.
6. 6a^ + 7a; = 160. 22. aj*^ ca? — + — = 0.
a a
23. a^2(a6)ajf &* = 2a6.
9. 2a^ + 3a. = 2. ^4. ___ = l.
10. 12ar* — » — 1 = 0. a a
25. ^ + ± = 3.
11. 0^ + 15 + 1 = 0. ^^ ^ + ^
^ ^ 2g _i 1 _ 3 + a?'
12. iB*44aj = 0. * 6 a; 6fa? ft^aj*
13 2^4.51 = ^"^^ Q^2ax + 12x = 24:a,
6 3 «« « . 2a 3
28. H = .
14. a^_?5 = 0. a 0. a
2 1 IS
29. ajhi = i?
15. a; = a4
« » 30. 6aj(aj3)=2^12a?.
16. _? + _iL.= 4. 31 3^±4__302^^7£~14,
a + x ax 5 a.__5 10
17. ba?(a\rab)x = a\ iq U2x 5
32. — — — = — •
IS. a?2aa = ba\ x a? 2
2S6 ALGEBRA. [§ 46{.
33. ?±ii£±I = 12z:^. 47. 2aj5VS = 3.
^ ^ ^^ 48. 2a;fl=V6a; + 3.
34. ?^ j —L = a? h 11. 49. aj 4 2 V3a; f 1 = 0.
5 aj — 4
60. ■y/a^^{Vb—x=
b
35. 10(x h 2) = ^^. ' '" \^J_I V6^
* 61. V5 + V5xf 1 = 2.
36. ^ + ^ = L 52. aj + 2V^ir5 = 5.
a — 2 a; — 3 2
37. 3(a?+4)242(a;17)*=274. ^^' Vx + 1+— ==2.
38. aj* + 4naj=:4a:* + l. o
39. a^a^ = ax. ^^' Va. + lhVS =
40. a^7aaj = 78al
41. ^H5±i5 = l ""' ~V^"^
a: + a a: 2
1 66. Vx — a^ ^
V« + l
55. ^±1 — 5±l.
42. aj = n. Va; + 2a
a;
I, 57. 2Va^64 = 2aj8.
43. aa;4 = c. , _
X 58. vaj + 5— Vx = l.
44. (m+w)ar^ 'HJlH.xz^m+n, 59. V2 a; + 4=^/5 4. 6 41
'^ m— n \2
45. a^n^ = m(2a:m). ^^ V2^+l= ^ + ^ +1,
46. c^ar^  acic = ^(6  a^). V2aj + 1
61. (5a;2)2(3aj + 2)« = (a;3)2l.
62. 1 1(^ + 3) , 4(0^^5) , 78
ajf4: aj4 a^16
63. (a^  6>2 _ 2(a2 4. 52)a; = 52 _ ^2^
64. V2a; + 9hV3a;15=V7ajh8.
65. 4=f ^ = ?.
a;+V2ar^ a;V2^ar^ ^
66. A/l + Vi = Vl42aj.
§470.] QUADRATIC EQUATIONS. 267
EQUATIONS IN THE QUADRATIC FORM.
468. Any equation is said to be quadratic in form when it
is composed of three terms, two of which contain the unknown
number with an exponent in one term twice its exponent in
the other. Thus, the equation 02;^ + 62?" = c is quadratic
in form.
469. Any equation reducible to the form 025*" + 62;" = c (1),
in which z represents any expression simple or compound, and
n any exponent, may be solved, at least in part, by the methods
for the solution of complete quadratics : for, denoting z* by
«, and hence z^ by a?, (1) becomes
aa? \hx = c'^
whence a. or ^ = ^±^J>' + ^<^.
2a
Whether the given equation can be completely solved will
depend upon the possible solution of the two equations
"^^ Y'a '
2f =
2a
470. The methods of procedure in solving such equations
will be best understood from the following examples.
1. Solve the equation aj* f 21 ar* = 100.
Treating 7? as the unknown number, we obtain
x^ =  lOJ ± V( V)2 + 100.
.. a;2 = _ioi ±i4i=4or 26.
Solving the equations,
aj2 = 4, a; = ± 2.
a;2 = _ 25, X = i 5 V^.
white's alo. — 17
268 ALGEBRA. t§ 470.
It is thus seen that the given equation has fonr roots, two being real,
and two imaginary; and it may be found by substituting that each
root will satisfy the given equation.
2. Solve the equation 3ofi + 20a? = 32.
Treating ofi as the unknown number, we obtain
^^ 10±Vi00T96 ^JL0^U^4^^_e
3 is 8
Extracting the cube root, x = V^ or — 2.
This equation has also four imaginary roots, the finding and explaining
of which belong to a more advanced treatise.
8. Solve the equation 5^/x — Ss/x = U,
This equation may be presented under the form
Treating x* as the unknown number, we obtain
x* = ^ ^ = 2 ^ 2 2 or — y
o 5 o
Hence x = (a;*)* = 2* or ( J)* = 16 or 3i^.
4. Solve the equation Va; + 4 — 5^a; + 4 = — 6.
Changing to (05 + 4)2 — 6 (as + 4)* = — 6, and treating (a; + 4)* as the
unknown number, we obtain
(x + 4)i = t ± V(i)2 6=±i = 3or2.
Hence a; + 4 = [(x + 4)*]* = 3* or 2* = 81 or 16.
/. a; = 77 or 12.
5. Solve the equation a?=zl.
Transposing, x* — 1 = ;
factoring, (a;  1) (x^ + a; + 1) = ;
equating factors to zero, x  1 = 0, .. x = 1 ; (1)
x2 ^. a; + 1 = 0. (2)
§472.] QUADRATIC EQUATIONS. 269
Solving (2) BS a quadratic, x = ^ — ^ "" —
Hence the equation x^ = l has three roots, 1, ^ — ^ — ^^^y and
— = — , one root being real, and the other two imaginary.
471. Certain equations which are not quadratic in form may
be put in the quadratic form by eliminating a factor.
Solve the equations
6. 0^100^ + 9 = 0. 19, a?aj* = 56.
20. a;fo=Va + o + 6.
8. aJ*6a^4 = 12. ^ ^^ , ^^^
« « , .o 21. a^20a» = 189.
9. a^2a^ = 48.
10. 0^4 a:« 28 = 4. ^2. a^7a^ +Va^7ar f 18=24.
11. o^ — 2po^=:q. 23. ic^+eVif^— 2aj+5=2ajll.
12. aj*3aj2 = 2VaJ*3a?. 24. 9aj+Var^3aj+5=3aj2+ll.
13. 7aj«12a^ + 5 = 0. 25. VS + S + ^t^^T3 = 6.
a?
14. a; 4 V5aj 4 10 = 8.
15. x+Vl0x\6 = 9. 26. 2(Vr=^+l)=
16. (a^ + 2)2 = 2aJ* + 8. n
i.y / — T^ . / — T^ o r 27. af — 2cKC2 = y.
17. Vaj + 6+V« + 3 = 3Va:.
18. Va + 2 = 2V2aj + l. 28. 3aj*aj* = 38.
VT+^1
29. Va* — a^ + 05 Va^ — 1 = a VI — aA
^o ^ ~ ^^ ax — o^ X
a + V2aaja^ a — «
472. Equations of the third and higher degrees may be
readily solved by factoring^ if, when the second member is
made zero, the first member is the product of rational factors.
Take, for example, the equation ic* — 7aj — 6 = 0,
260 ALGEBRA. [§ 472.
Factoring by synthetic division (§ 215),
equating factors to zero, aj + l=0, ajh2 = 0, a? — 3 = 0;
whence x = — l, 05 = — 2, x = 3.
Hence the roots are — 1, — 2, and 3.
Solve by factoring by synthetic division
31. o^lOaj* ♦31a; 30 = 0. 33. a*5a^f4 = 0.
32. a^3i?Sx + 12 = 0. 34. aJ*4«»7a^f 34aj24=0.
PROBLEMS INVOLVING QUADRATICS.
1. The perimeter of a rectangular field is 600 yards, and its
area 14,400 square yards. What is the length of the sides ?
600 yd. T 2 = 250 yd., length of two adjacent sides.
Let X = length of one side ;
then 250 — x = length of the other side.
Hence, by the conditions,
x(250 x)= 14400 ;
whence x^250x =  14400.
Solving equation, « = 125 ± 35 = 160 or 90.
If we take x = 160, the other side is 90, and vice versa. Thus, though
X has two values, the problem has but one solution.
2. By selling a lot of goods for $ 24, a merchant lost as
many cents on the dollar as he paid dollars for the goods.
How much did he pay?
Let X = number of dollars paid ;
then a; — 24 = number of dollars lost.
Hence, by the conditions, x x x = (x — 24) 100.
Simplifying, etc., x^ _ iqo x =  2400 ;
solving equation, x = 60 or 40.
Either of these values of x satisfies the conditions of the problem : for,
if he paid $60 for the goods, he lost 60% by selling them for $24 ; and, il
he paid $ 40 for the goods, he lost 40 % by selling them for 1 24. It is thus
seen that the problem admits of two solutions.
§473.] QUADRATIC EQUATIONS. 261
3. A drover bought a number of oxen for $ 400; but, if he
had obtained 4 more for the same money, the price paid per
head would have been f 5 less. How many oxen did he buy ?
Let X = number bought ;
then 452 = price per head,
X
400
and = price per head if 4 more had been bought.
05+4
Hence, by the conditions,
400 ^5^400
x + 4 X
Solving equation, « = 16 or — 20.
The negative root (—20) is not admissible, since it will not satisfy
the conditions of the problem, and hence the number of oxen bought
was 16.
4. The sum of the ages of a father and son is 65 years, and
the product of their ages is 250 more than 10 times their sum.
What is the age of each ?
Let X = father's age ;
then 66 — a; = son's age.
Hence (65  x) x = 900.
Solving equation, x = 45 or 20.
The father's age is 45 years, and the son's 20 years. Here the second
value of X is inadmissible, since it would make the son older than his
father.
6. Divide the number 12 into two parts such that their
product will be 40.
Let X = one part ;
then 12 — X = the other part
Hence (12  x) x = 40.
Solving the equation, x = 6 ± 2V— 1.
Both values of x are imaginary, and hence the given problem is
impossible.
473. It is shown by the above examples that the solution of
problems involving quadratics may give results all of which do
not satisfy the conditions of the problem. This is due to the
262 ALGEBRA. [§ 474.
fact that there may be limitations in the problem, expressed or
implied, which do not appear in the eq^uation.
Hence, in solving problems, only those values should be
retained as answers which satisfy all the conditions of the
given problem. The arithmetical values that will satisfy a
problem are posUivef and usually the first values found.
474. In some cases a change in the wording of a given
problem will form an analogous problem, to which the absolute
value of the negative root found is an answer. Thus, if Prob
lem 3 above be so changed as to state, that, if 4 less oxen had
been bought, each would have cost $ 5 more, the values of x
would be 20 and — 16, the latter being inadmissible.
Imaginary roots indicate that the problem is impossible.
6. The sum of two numbers is 17, and their product 60.
What are the numbers?
7. The sum of two numbers is 72, and their product is 10
times as great. What are the numbers ?
8. The sum of two numbers is 50, and the sum of their
squares is 1282. Find the numbers.
9. The difference of two numbers is 7, and their product is
2340. Find the numbers.
10. The sum of two numbers is m, and their product is n.
What are the numbers?
11. The sum of two numbers is m, and the sum of their
reciprocals is n. What are the numbers?
12. Find a number such that 5 times its square increased by
10 times the number itself will equal 495.
13. Find three numbers such that the second will be one
half of the first, and the third one third of the first, and the
sum of their squares will be 441.
§474.] QUADRATIC EQUATIONS. 263
14. A certain number of persons pay together a bill of
$ 190, each paying f 9 less than the number of persons. How
many were there, and how much did each pay ?
15. A piece of groimd one rod longer than broad contains
1190 square rods. What is the length ?
16. An army corps consisting of 12,850 soldiers was formed
into two squares, one of which had 10 more men in a side
than the other. How many men in each square ?
17. A man made a journey of 48 miles in a certain number
of hours. If he had traveled 4 miles more per hour, he would
have made the journey in 6 hours less time. H!ow many miles
per hour did he travel ?
18. A farmer bought a number of sheep for $ 80. If he had
bought 4 more sheep for the same money, he would have paid
$ 1 less per head. How many sheep did he buy ?
19. A man, being asked his age, replied, "If to the square
root of my age you add ^ of my age, the sum will be 26 years."
What was his age ?
20. Two couriers, A and B, start at the same time to go to
a place 90 miles distant. A traveled 1 mile per hour faster
than B, and reached the place 1 hour before B. At what rate
did each travel ?
21. A merchant bought a number of fur robes for $150,
and then sold them at $ 18 a robe, and thus gained on each
robe twice the cost of a robe. How many robes did be buy ?
22. A farmer has two square fields. The side of one is 2^
rods longer than the side of the other, and the fields together
contain 1131^ square rods. How many more square rods in
the larger field than in the smaller ?
23. A man sold a farm for f 3150, and afterwards bought
another farm containing 7 more acres for the same money, at
$ 5 less per acre. How many acres in the farm sold, and how
much did he receive per acre ?
264 ALGEBRA. [§ 474.
24. A certain number of pieces of cloth cost $ 1260, each
piece costing $9 more than 5 times the number of pieces.
How many pieces were there^ and how much did each cost ?
25. A certain number of persons equally engaged in a
business transaction lost $96,000; but, four of them becoming
insolvent, each of the rest had to pay 1(4000 more than his
fair share. How many persons were engaged in the business ?
26. A jeweler sold a watch for $ 144, gaining on it as much
per cent as the watch had cost him. How much did it cost
him?
27. A number of horses were bought for $ 1800. Had 3
more been obtained for the same money, each would have cost
$ 30 less. How many horses were bought ?
28. A cistern supplied by two pipes could be filled by one
alone in 5 hours less than by the other alone, and both
together could fill it in 6 hours. In how many hours could
each fill it alone ?
29. A steamer performed its down trip of 150 miles at a
certain rate per hour. On the return trip, going 3 miles an
hour slower, it took 2^ hours longer. What was the rate
down the river ?
30. A man bought a certain amount of sugar for $ 66 ; but,
if sugar were to rise one cent per pound, he would obtain 50
pounds less for the same money. How much sugar did he buy ?
31. A drover bought a certain number of sheep for $483.
Reserving 20 of the number, he sold the rest for $ 432, gaining
1 on each. How many sheep did he buy ?
32. What is the price of eggs per dozen when 3 less in 25
cents' worth raised the price 5 cents per dozen ?
Find a general formula for the above problem, putting a, 6,
c, for the above numbers.
§477.] SIMULTANEOUS QUADRATIC EQUATIONS. 266
CHAPTER XVI.
SIBIULTANEOUS QUADRATIC EQUATIONS.
476. The solution of simultaneous equations of the second
degree with two unknown numbers involves the elimination of
one of the unknown numbers, and the forming of an equation
with but one unknown number. This may lead to an equation
of a higher degree than the second, usually of the fourth,
which cannot in general be solved by quadratic methods.
476. There are, however, several cases in which the solution
of such simultaneous equations can be effected by equations of
the second degree. The three cases of most frequent occur
rence are
I. When one of the given equations is simple. .
II. When the equations are homogeneous and quadratic,
III. When the equations are symmetrical with respect to the
unknown numbers,
A few solutions will sufficiently illustrate the process in each
case.
477. I. One of the given equations simple.
1. Solve the equations \^ + 3xyf^2% (1)
^ ( 4.xy = 7, (2)
From (2), transposing, y = 4 x — 7 ; (3)
substituting value of y in (1), x^ + 3«(4 «  7)  (4 «  7)2 = 29 ;
whence 3 x^  35 x =  78. (4)
Solving (4), x = 8or3;
substituting value of x in (2), y = 27 or 6 ;
.«. X = 8, y = 27 ; or x = 3, y = 6.
266 ALGEBRA. [§ 47a
If preferred, the values of x and y may be braced in corresponding
naiia thus • / * = ^' ^ = ^»
pairs, thus. ^^g^^^^27§.
478. In like manner simultaneous quadratic equations^ when
one is simple, may be solved by finding in the simple equation
the value of one of the unknown numbers in terms of the other,
and substituting in tfie other equation.
Solve the following groups of equations :
2 <x» + 3/' = 89, g (5x^23^ = 93,
\xy = 3. ' \3x~4y = l.
3 <x'f=16, ^ ia^ + 2f = 9,
' \2x\ry = 13. ' }.x + 2y = 5.
\
ar^7y2 = l, f5aj«3a^ + 2^ = 45,
* x~2y = 2. ' \Sx^2y = 22,
6 (a^2/= = 9, 3 \10x + y=3xy,
C2a; + 2/ = 14. * \y — x = 2.
10.
11.
7/2 + 20^=11,
f 3a; = 9.
if
\2y
( 5x2 _Sxy + y^ = 2xSy^ 31,
\5x2y = ll,
^^ Ux2y + (y6y = 692xy,
(5aj — 4y = l.
479. II. The equations homogeneous and quadratic.
An equation is said to be homogeneous when all its terms
which contain the unknown number are homogeneous (§ 56),
13. Solve the equations I f + ^ f ^^' (^)
^ \2xyf=:3. (2)
Multiplying the first member of (1) by the second member of (2), and
the first member of (2) by the second member of (1), we have
Sx^ + Zxy = 20xy  lOy^ ; (8)
or 3x2 _ 17 yx =  10y2. (4j
§481.] SIMULTANEOUS QUADRATIC EQUATIONS. 267
Considering 17 y as the coefficient of x in (4), and solving the equation
as a quadratic, we have
fl5 = 5yorf y;
whence 2 acy = 10 y^ or J y^.
Substituting values of 2icy in (2), 9y2 = 3 ; Jy^ _ ^2 _ j^a _ 3 .
whence y = ± iVS; y =±S,
.. x=±iVS; x=±2.
In like manner any two homogeneous equations of the second degree
may be solved.
480. Two homogeneous simultaneous equations of the second
degree may also be solved by assuming y = vx, and substitut
ing vx for y in both equations, and then by division or other
wise obtaining an equation involving only v. Having found
the value of v in this equation, the values of x and y may be
found by substitution. It is, however, believed that the method
illustrated by Example 13 is simpler.
Solve the following groups of equations :
(4 0^3 0^ = 18. * lxy = 6.
{2fxy = 104.a^, <2xy ^24. = Sx',
l2x'^3xy = 2f. ' (/_a^ = 3.
<2f^xy = 5, <a^2xy==9^Sfy
lx'2xy = f\2. • (3^40:3^ = 563^.
^^ (a^23^ = 8, 22 P^3a:3^ = 43^,
l3fxy = 4:. ' \a^2xy = 93f.
18. a^+2^ = 6i, 23. \<^y)+y(^y)=^^s,
ia^ — xy = 6. ' (7 x(x+y)=72y(x — y).
481. III. The equations symmetrical.
An equation is said to be symmetrical with respect to two
numbers, as x and y, if the numbers can be interchanged
without destroying the equality.
16.
268 ALGEBRA. [§ 482.
Thus, a? — 2{cy + j^ = 12ia symmetrical ; for, on interchang
ing X and y, it becomes jf — 2yx + a^ = 12, a true equation,
since x^ — 2xy\y^ = y^ — 2yx\a^ (§ 80). The equation
0?^ — ^ = 16 is not symmetrical, since aj* — ^ is not equivalent
to y* — aj*.
24. Solve the equation \^'^^^7^^' ^^^
^ U + y = 12. (2)
Squaring (2) , x^ +2xy + y^ = 1U; (3)
subtracting (1) from (3), 2xy = 70; (4)
subtracting (4) from (1), x^ — 2xy {y^ = ^;
extracting square root, x — y=±2; (6)
2 X = 14, . •. x = 7 ;
10, . •• » = 5.
Substituting 7 for a; in (2), y = 6 ;
substituting 6 for « in (2), y = 7.
Hence the values of x and y are x = 7, y = 6; x = 6, y = 7,
adding (5) and (2), ^^^
482. In like manner any two simultaneous symmetrical
equations of the second degree may be solved by so combining
them as to obtain the values of the sum and the difference of the
unknown numbers,
483. Groups of equations which are symmetrical except in
the signs of the terms may often be solved by the symmetrical
method.
25. Solve the equations < ^ ' )r,i
(xy = '2. (2)
Squaring (2), x^2xy+y^ = i] (3)
subtracting (3) from (1 ) , 2 icy = 48 ; (4)
adding (1) and (4), ofi \ 2 xy \ y^ = 100 ;
extracting square root, x{y = ±10; (6)
adding (2) and (6), 2 a; = 8 or  12, .*. a; = 4 or  6 ;
subtracting (2) from (6), 2 y = 12 or  8, /. y = 6 or — 4.
.*. X = 4, y = 6 ; or ac = — 6, y = — 4.
§485.] SIMULTANEOUS QUADRATIC EQUATIONS. 269
26. Solve the equations ^ + 2^ = ^^^ 0)
(x + y = 5. (2)
Dividing (1) by (2), x^xy + y^ = 7; (3)
squaring both members of (2), x^ \ 2 xy \ y^ = 2b ; (4)
subtracting (3) from (4), 3 a;y = 18 ;
whence ojy = 6. (5)
Subtracting (5) from (3), 7^2xy + y^=l;
extracting square root, « — y = ± 1 ; (6)
adding (2) and (6), 2 a; = 6 or 4, /. ac = 3 or 2.
Substituting 3 or 2 for a; in (2), yz=z2 or 3.
484. Groups of symmetrical equations often fall under the
first case, and are readily solved by substitution.
Solve the following groups of equations :
\xy = 35, \x\y = 7.
\xy = — 15. \x — y = 3,
29. \<^+f = Qh 36. fa^«^ + 2^=19,
(^ + y = — 1. ix — y = S,
30. «' + 2' = 5, 3g a^3/« = 37,
( a!*y + xjf* = 30. {x — y = l.
32, (0^ + 2/^ = 29, 33^ j a^ + xy ]f = 175,
27.
28.
{
Uy = 3. la^2^ = 875.
SPECIAL METHODS.
485. The preceding methods of solving simultaneous equa
tions of the second degree are called general because they
apply to classes of equations. There are, however, many
simultaneous equations not falling under these cases, which
are readily solved by special artifices. Some equations that
do fall under these cases may be solved more elegantly by
special methods.
270 ALGEBRA. [§ 485.
1. Solve the equations i ^i /2_aa
a^  y* = 369, (1)
(2)
Dividing (1) by (2) , x^y^ = 9; (3)
adding (2) and (3), 2 x* = 50 ; .. jb = ± 6.
Subtracting (3) from (2), 2 y^ = 32 ; .. y = ± 4.
Hence a! = 6, y = 4; orx=— 5, y= — 4.
This solution gives only the finite roots of equations (1) and (2).
2. Solve the equations «* + »* = ^'^^ 0)
^ I x\y = 5. (2)
Raising (2) to the 4th power, we have
«* + 4 ie8y + 6 a:ay2 + 4 acy* + y* = 626 ; (3)
adding (1) and (3), and dividing by 2,
«* + 2 a% + 3a;V + 2 xy3 + y* = 361 ;
extracting the square root, x^ + xy + y^ = ± 19. (4)
We now have two equations, (2) and (4) , which can be readily solved
by substitution or by the symmetrical method.
1
3. Solve the equations
a2 + 2a^ + i» + 3y = 76, (1)
f + x^y = 24, (2)
Adding (1) and (2), x^\2xy + y^ + 2x^ 2y = 99; (3)
factoring parts and adding 1, (x + y)^ + 2(a: + y) + 1 = 100 ;
extracting square root, x + y + 1 = ± 10.
/. X + y = 9 or  11. (4)
Equations (2) and (4) can now be solved by substitution, the first
general method.
Solve the following groups of equations :
^ (a^^f^l6, ^ <a^ + f = m,
\x + y = S, ' \x + y=2l2,
(x^f=.S2, (0^2^ = 544,
Xxy = 2, ' la^ + 2/* = 34.
(a^^f = 19, (aj*y*=1280,
• Uy = l. * la^f = 32.
§486.] SIMULTANEOUS QUADRATIC EQUATIONS. 271
a^4y* = 706, ^ (aj* + 2/* = 272,
10.
11.
12.
13.
faj* + y* = 706, ^^ (aj* + 2/* = 2
\x\'y = S. ' \x — y=:2.
(a^^3xy + f = B9, (a^ + y* = 2657,
W + 2^ = 29. * U + y = ll.
a^ + l = a? + y, ^g (aj« + 2ajy + y + 3aj = 73,
j«* + y* + « + y = 18, j^^ <oi^ + xy = a^ + ab,
\xy = 6. ' \ j^ + yx = b^ \ (ib.
PROBLEMS INVOLVING SIMULTANEOUS QUADRATICS.
486. The statement of problems involving two or more un
known numbers has been sufficiently illustrated under the head
of simple equations. Several of the problems below can be
solved by the use of only one unknown number, but their
solution is facilitated by the use of two.
1. The difference of two numbers is 7, and the difference
of their squares is 119. What are the numbers?
2. The sum of two numbers is 18, and the sum of their
squares is 170. What are the numbers?
3. The sum of two numbers is 25, and the difference of
their squares is 175. What are the numbers?
4. Find two numbers such that the first increased by twice
the second is 24, and the sum of their squares is 149.
5. The sum of the squares of two numbers exceeds twice
their product by 9, and the difference of their squares is 1
less than their product. Find the numbers.
6. The sum of six times the greater of two numbers and
five times the less is 50, and their product is 20. Find the
numbers.
7. The product of two numbers diminished by their sum
is 17, and the sum of their squares is 65. Find the numbers.
272 ALGEBRA. [§ 48«.
8. Find two numbers such that their sum is 19^ and the
sum of their cubes 1843.
9. Find two numbers such that their difference is 4, and
the difference of their cubes 988.
10. The product of two numbers multiplied by their sum
is 180, and the sum of their cubes is 189. Find the numbers.
11. If a certain number expressed by two digits be multi
plied by the sum of its digits, the product will be 160 ; and, if
the number be divided by four times its unit digit, the quo
tient will be 4. Find the number.
12. What number divided by the product of its two digits
is 5^, but, when 9 is subtracted from it, the resulting number
is expressed by the two digits in an inverse order ?
13. The area of a rectangular field is 300 square rods, and
the length of its diagonal is 25 rods. Find the length of the
sides.
14. The sum of the diagonal and the longer side of a
rectangle is three times the length of the shorter side, and
the difference in the lengths of the two sides is 4 yards. What
is the area of the rectangle ?
15. The area of the floor of a certain hall is 5375 sq. ft., and
its length is 4 feet less than three times its breadth. What
are the dimensions of the floor ?
16. A certain number of sheep were bought for $ 468 ; but,
after 8 of them had been reserved, th« rest were sold at an
advance of f 1 a head, and $ 12 were gained on the lot. How
many sheep were bought ?
17. A vessel can be filled in 6 hours by two pipes running
at the same time, but one pipe can fill it alone in 5 hours less
than the other. How many hours does each pipe require to
fill it?
§ 491.] INEQUALITIES. 273
CHAPTER XVII.
DTEQUALITIES.
487. The expression a>h denotes that a is greater than 6,
and a<b denotes that a is less than b (§ 39). The sign > or <
is called the sign of inequality.
488. An inequality is an expression consisting of two unequal
numbers connected by the sign of inequality. Thus 4 > 3 and
a? < y are inequalities.
489. Two inequalities are said to subsist in the same sense
when their first members are both greater or both less than
their second members. Thus, a > 6 and c > d subsist in the
same sense.
490. Two inequalities are said to subsist in a contrary sense
when the first member is the greater in one, and the less in the
other. Thus, a > 6 and c < d subsist in a contrary sense.
Inequalities are also called inequations; and two inequalities which
subsist in the same sense are also said to be of the same direction, since
the signs point in the same direction ; and two inequalities that subsist in
a contrary sense are also said to be the reverse, since the signs point
iu opposite directions.
It is assumed in this chapter, unless the contrary be stated, that the
letters denote real and positive numbers.
491. 5 > 3, and 5  2 > 3 h 2 ; and, generally, if a > 6, then
a\c>h{c. Likewise 5 > 3, and 5 — 2 > 3 — 2 ; and, gener
ally, if a > 6, then a — c'>h — c. Hence, if the same positive
number be added to or subtracted from both membei^s of an in"
equality^ the resulting inequality will subsist in the same sense.
WHJTJ£*S JLLQ, 18
274 ALGEBRA. [§ 492.
492. It follows that a term can be transposed from one
member of an inequality to the other, as in an equation, pro
vided its sign he changed,
498. If a>6, then 2a>26, 3a>36, and ac>6c; and, if
a > 6, then  > ,  > , and  > — Hence, if both members of
2233 c c
an ineqvxdity be multiplied or divided by the same positive num^
ber, the resulting inequality will subsist in the same sense,
494. A positive number is greater, algebraically considered,
than any negative number ; and, of two unequal negative num
bers, the less numerically considered is the greater algebraically
(§ 67). Thus, 2 > 7, and  2 >  7.
Hence, if both members of an inequality be multiplied or
divided by the same negative number, the resulting inequality will
subsist in a contrary sense. Thus, 2<,5, but 2x(— 3)>5x
(3); also 2>5, but 2 x(3)<5 x(3).
495. It follows, that, if the signs of both members of an in
equality be changed^ the resulting inequality wiU subsist in a
contrary sense.
496. If a>by c>dy and e >f then a\c\e>b\d +/,
and axcxe^bxdxf Hence, if two or more inequalities
that subsist in the same sense be added member to member , or
multiplied member by member, the resulting inequalUy wHl subsist
in the same sense,
497. If an inequality be subtracted from, or divided by,
another inequality in the same sense, the resulting inequality
may or may not subsist in the same sense. Thus, 5 <S, and
1 < 5, but 5  1 > 8  5; 4 < 6, and 1 < 3, but 4 f 1 > 6 ^ 3.
These operations are to be avoided when the sense of the
resulting inequality cannot be determined.
498. 'if a > b, then a^ > b\ a^ > l^, and a" > b\ Hence, if
both members of an inequality of positive numbers be raised to the
same power, the resulting inequality will subsist in the same sense.
§ 501.] INEQUALITIES. 276
499. If the members of an inequality are negative^ the
same odd powers will subsist in the same sense, and the same
even powers in a contrary sense. Thus, if — a > — 6, then
(__ of > ( 6)^ but ( of < ( h)\
500. If a* > 6*, then a^ > W and a > 6. Hence, if the same
root of both members of an inequality of positive nurabers be taken,
the resulting inequality will subsist in the same sense.
601. An inequality, like an equation, may be simplified by
certain transformations ; and an inequality is said to be solved
^when a limit to the value of the unknown number is found.
1 . Simplify the inequality ^  2 > 3 + ^.
Transposing  2 and ^, ^  ^>5;
clearing of fractions, and uniting, 5 a; > 30 ;
dividing by 6, oc > 6.
Hence 6 is a limit of the value of x.
2. Simplify 2a! + 4>^.
3. Find the limits of x, when given <
!+»<!+„ (1)
Transposing in (1), ^  < a  6 ;
b a
clearing of fractions, <mc — 6x < a^ft — a6^ ;
factoring, (a — h)x <(o — h)ah ;
dividing by a — 6, x < a6.
Clearing (2) of fractions, ox — 6a; > o* — o5 ;
factoring and dividing by a — 6, x > a.
Hence the limits of x are a and ah.
4. Find the limit of aj in — < "" %
b X b»
276 ALGEBRA. [§ 502.
Find the limits of a; in
5.
(10x<3x + 49.
X a — b
8. <
6. <
ich5> + 55.
2 3 ^2"~ ^^2""
•1
a — 6 a:
ax — ox<i .
X
aa \ bx > ab + h^,
fAx2 2Ax
"^ X . ^x
7. <
3
a
3x2<^ + ^. ^*' )? + ?>« + '»
2 o 6 a
\i
502. TA6 »t^m q/* the squares of any two unequal numbers
is greater than twice the product of the numbers.
For let a and b be any two unequal numbers. Then, since
(a — by is positive whether a> or < 6, we have
(aby>0.
Expanding, a* — 2a66*>0;
transposing —2ab, a^ + 6^ > 2 ab.
11. Show that the sum of any fraction whose terms are
unequal and its reciprocal is greater than 2.
Let  be any fraction in which a > or < 6. Then, by § 502, we may
b
assume that o^ + 6^ > 2 ab.
Dividing by a6, ^ + >2.
b a
Assume that a and b are positive unequal numbers, and
show that
12. a^b + ab^>2aV. ,. a±b^ 2ab
14. — — — ^
13. a« + 6«>a26 + a6l 2 a + 6
§ 506.] RATIO. 277
CHAPTER XVIII.
RATIO, PROPORTION, VARIATION.
RATIO.
503. Ratio is the relation of one number to another of the
same kind expressed by their quotient. Thus^ the ratio of
a to 6 is •
Every fraction expresses the ratio of its numerator to its denominator ;
and every integer expresses the ratio of itself to unity.
604. A ratio may be expressed by writing a colon (:) be
tween its two terms. Thus, the ratio of a to 6 is expressed
by a : 6, read " the ratio of a to 6," or, briefly, " a to 6."
The ratio of a to 6 is expressed by a : 6, and the ratio of 6 to a by 6 : o ;
but the ratio between a and 6 or 6 and a is expressed hy a:b or bict.
Thus, the ratio of 3 to 6 is f, and the ratio of 6 to 3 is } ; but the ratio
between 3 and &^r 6 and 3 is  or f .
505. The first term of a ratio is called the antecedent; and
the second term, the consequent. The two terms of a ratio
taken together are called a couplet.
The antecedent of a ratio is the dividend ; and the consequent, the
divisor.
506. When the antecedent equals the consequent, as a: a,
the ratio equals unity, and is called the ratio of equality.
When the antecedent is greater than the consequent, as 8 : 5,
the ratio is greater than unity, and is called a ratio of greater
inequality.
278 ALGEBRA. [§ 507.
When the antecedent is less than the consequent, as 5:8,
the ratio is less than unity, and is called a ratio of less
inequality.
When the antecedent and the consequent are interchanged,
the resulting ratio is the inverse of the given ratio. Thus,
6 : a is the inverse of a : 6.
607. Since 2 = 5L2i^, and 2 = ?L±«, hoth terms of a ratio
b b xn b 65n
may be multiplied or divided by the same number without aUer
ing the value of the ratio.
506. If  = r, " ^ ^ = m, and — ?L.= rii; hence multiply
b b b sn
ing the antecedent or dividing the consequent of a ratio by a
number multiplies the ratio by that number.
509. If  = r, ^^^ or —55— = '^ ; hence dividing the ante
b b b xn n
cedent or multiplying the consequent of a ratio by a number
divides the ratio by that number.
510. Eatios may be compared by reducing the fractions that
express them to a common denominatory and comparing the result
ing numerators.
a o
Thus, to compare a : b with c:d, we reduce  and  to
ad be b d
— and — respectively, and then decide that a:b > or =
bd bd
or < c : d, according as ad > or = or < be.
511. If a>b, ^L±^<«; and if a<b, ^>?.
b {n b b {n b
For, reducing the ratios to a common denominator, we have
a^n _ ab{bn •, a _ db \ an
b + n^bip + ny b~ b(b\n)
If a>6, (ab + bn)<i(ab + an), and hence 5L±_5<?; but, if
b \ n b
a<b, (ab h bn)>(ab {■ an), and hence ^"^^ >~.
6 f n b
§ 515.] RATIO. 279
Hence a ratio of greater inequality is diminished, and a ratio
of less inequality increased, by addiiig the same number to both
terms of the raJbio,
512. If a > 6, ; > • ; and, if a < ft, ; < •
b —n b b — n b
For, reducing the ratios to a common denominator, we have
a — n _ ab — bn j a _ ab — an
b — n b(b — n) b b(b — n)
If a>b, (ab — bn) > (ab — an), and hence "~ >  ; but, if
a<b, (ab — bn) < (ab — an), and hence ~" < •
b — n b
Hence a ratio of greater inequality is increased, and a ratio
of less inequality diminished, by subtracting the same nuraber
from both its terms.
The principles stated in §§511, 512, may be thus illustrated :
(1) ±<4 ^^d ^>?; (2) ^^>K and ?:i2<?.
^^3 + 2 3 4 + 2 4 ^^32 3 42 4
613. Generally, since every ratio may be expressed as a
fraction, whatever operations on the terms of a fraction affect its
value, will in like manner affect the value of the corresponding
ratio,
514. A compound ratio is the product of two or more ratios.
Thus, the ratio axiibd is the product of a : 6 and c : d, and is
called compound.
Two or more ratios may be compounded by taking the product
of the fractions that express them. For example, the ratios
oi a:b, c:d, and e :f, are compounded by taking the product of
, , and —• Thus, r X  X — =
ace
b' d' f ' b d f bdf
615. A ratio, as a : 6, may be compounded with itself, a^ : &'
280 ALGEBRA. [§ 516.
is called the dujf^iccUe of the ratio a : 6 ; cfib^, the triplicate of
aib'j and so on.
The ratio Va : y/b or a^ : b^ is called the subduplicate of
a:b; and v^a : ^6 or a^ : 6*, the subtriplicate of a : 6.
516. When one or both terms of a ratio are incommensurable
(§ 371), the ratio is said to be iticommensurable. Thus, 1 : V2
is an incommensurable ratio.
Problbhs.
1. What is the ratio of 2 lb. to 2 oz. ? Of f 0.75 to $ 3 ?
2. Arrange in descending order of magnitude
4:5, 7:3, 12 : 4, 3:8, 5 : 12, 7:5.
3. What is the ratio compounded of 2 : 3 and 15 : 16 ? Of
7 : 6 and 24 : 35 ?
4. Two numbers are in the ratio of 3 to 5; but, if each
be increased by unity, their ratio becomes 11 : 18. Find the
numbers.
5. Two numbers are in the ratio of 8 to 7; but, if each
be increased by 6, their ratio becomes 6 : 5. Find the num
bers.
6. Give the duplicate and triplicate ratios of 5 : 7, the
subduplicate ratio of 81 : 25, and the subtriplicate ratio of
729 : 343.
7. Reduce the following ratios to their lowest terms :
63:45, 138:124, a": ax, rrv"  vJ" : m^ + n\
8. Find two numbers in the ratio of 4 to 5, such that their
difference is to the difference of their squares as 1 to 27.
9. Find two numbers such that the ratio of their sum to
the sum of their squares will be as 11 to 195, the ratio of the
numbers themselves being 4 : 7.
10 o Find X so that the ratio of a? to 1 may be the duplicate
of the ratio of 8 to x.
§ 521.] PROPORTION. 281
PROPORTION.
617. A proportion is the expressed equality of two ratios.
This equality may be expressed by the sign : : or the sign =.
Thus, the equality of the two ratios a : b and c : d may be
expressed by a:b::c:d or by a:b = c:d, each being read
" the ratio of a to 6 equals the ratio of c to d," or, briefly, " a is
to 6 as c is to d"
518. Since each ratio has two terms (an antecedent and
a consequent), a proportion necessarily consists of four terms,
the first and third being antecedents, and the second and
fourth, consequents.
The first and fourth terms of a proportion are called the
extremes; and the second and third, the means. Thus, in
a:b::c:dy a and d are the extremes ; and b and c, the means.
619. Four numbers are said to be proportional, or in pro
portion, when the ratio of the first to the second equals the
ratio of the third to the fourth. Thus, a, b, c, d, are propor
tional when a:b = c: d. The four terms of a proportion are
called proportionals, the fourth term being called the fourth
proportional.
620. Numbers are in continued proportion when the ratios
of the first to the second, the second to the third, the third
to the fourth, etc., are equal. Thus, a, b, c, d, e, etc., are in con
tinued proportion when a:b = b :c = c:d = d:e, etc.
621. Since the ratio a : b may be expressed by the fraction
, and c\d by , the proportion a\b = c:d is identical with
b d
the equation  = ; and, by simple transformations of this
b d
fundamental equation, the following propositions in proportion
are proved.
282 ALGEBRA. [§ 522.
Propositions.
522. If four numbers are in proportion, the product of the
extretnes equals the product of the means.
Let
a:b = c:d'j
then
a_c
b~d
Multiplying by 6d,
cul = bc.
523. It follows from the above, that, if any three terms of a
proportion are giveiiy the other term can be found.
For, if ad=bcy then a = — , d = — : b = — , c = — : and
da c b
hence either extreme of a proportion may be found by dividing
the product of the means by the other eoctreme; and either mean
may be found by dividing the product of the extremes by the otiier
m^an.
Find the value of x in the following proportions :
1.
5:8
: : 15 : 0?.
6.
x:10+x::6:9.
2.
x:5 ,
: 6 : 10.
6.
aj : 15  a? : : 30 : 15.
3.
5:x\
: 7 : 10.
7.
6 : « : : 24 : aj + 6.
4.
4:6
: : a? : 4.
8.
4 : 18 : : a  3 : a? + 4.
5S4. If the product of two numbers is equal to the product of
two other numbers, the four nunibers are proportionals ; and any
twi) of them may be made the extremes, and the other two the
means, of a proportion.
Let ad = 6c.
Dividing by M, od^bc
^ ^ ' bd bd'
whence r = ; i.e., a:6 = c:<t
d
c
'd'
b__
d.
a
c '
b_
d
c '
§ 527.] PROPORTION. 288
The equation ad = &c, if divided successively by cd, ac, and
dc, gives
a b ' y jt
 =  ; i.e., a : c = : a ;
i.e., b:a = d:c;
— =  ; i.e., b: d=:a:c.
d c
Let the pupil obtain these proportions from ad = bc, and
compare them with the next two propositions.
Let the pupil illustrate the above, and also each of the fol
lowing eight propositions, by means of numbers.
625. If four numbers are in proportion, they will be in propor
tion taken alternately ; i.e., the first term will be to the third as the
second term to the fourth.
Let a\b = c\d\
then _ = _.
b d
Multiplying by , "" ~ 3 » ^®*' ^ • ^ = & : d.
C C Cv
526. If four numbers are in proportion, they will be in propor
tion taken inversely; i.e,, the second term wiU be to the first as the
fourth term to the third.
Let a:b = c:d;
then 2 = i.
b d
Dividing 1 by each member, 1 i  = 1 5  ;
b d
whence  =  > !•©•> b:a = d:c.
a c
627. When a h 6 is to 6 as c + d is to d, the numbers a, b, c,
and d are said to be in proportion by composition.
284 ALGEBRA. [§ 52a
When a^b is tobssc — d is to dy the numbers (hb,Cy and d
are said to be in proportion by division.
When a^b istoa — 6 as c + distoc — d, the numbers
a, by Cf and d are said to be in proportion by composition and
division.
628. If four numbers are in proportion^ they are in proportion
by composition or division.
Let a'.b = c:d\
then . 55 = f.
b d
Adding ± 1 to each member, 7 ± 1 =  ± 1 ;
b d
whence a±b^c_±d
b d '
that is, a±b\b = c±d'.d.
529. If four numbers are in proportion^ they are in proportion
by composition and division.
Let a:b = c:d\
then (§ 528) ^^ = ^±^, (1)
b d
and 2l=±^^.^=1. (2)
b d ^ ^
Dividing (1) by (2), member by member,
— 6 c — d^
that is, af^a — & = c4d:c — d.
530. In a series of equal ratios, the sum of the antecedents is tc
the sum of the consequents as any antecedent is to its consequent.
then, by § 522, ab = ba, ad = be. af= be, ah = bg.
Adding and factoring, a{b \ d +/4 h^= b(a + c + e 4 g') ;
that is, ahcf^f5r:6 + d +/+ * = a : 6.
or
§ 534.] PROPORTION. 285
631. The pivduct of the corresponding terms of two or more
proportions are in proportion.
Let a:b = c:dy
and e:f=g:hy
a c ji e g
b d f h
Multiplying member by member, tt = „ >
of dh
that is, ae:bf=cg: dh.
532. If a, b, c, d, are in continued proportion (§ 620), a is to
c as a* 18 to b*, and a is to d cw a^ is to h\
Let a:b = b:c = c:d,
a b c
6 c d'
whence 2x = x;
b c b b
a a * 9 1.Q
or =^; ie.; a:c = a^:Cr,
c W
b c d b b b
CL fit
or  = i: ; i.e., a:d = a^:b^.
d b^' '
533. When three numbers, a, 6, c, are in continued propor
tion, b is called a mean proportional to a and c, and c a third
proportional to a and 5.
534. The mean proportional to two numbers is equal to the
square root of their product.
Let a : 6 = 6 : c ;
then r = '
b c
Clearing of fractions, b^ = a>c;
whence b = Vac.
286 ALGEBRA. . [§535.
535. Several of the foregoing theorems are useful in the solu
tion of certain numerical problems, and also certain equations.
536. Prove the following propositions, and illustrate each
with numbers :
I. If two proportions have the same couplet in each, the
other couplets will form a proportion.
II. If two proportions have the same antecedents, the con
sequents are in proportion.
III. If three numbers are in proportion, the ratio of the first
to the third is the duplicate ratio of the first to the second.
IV. If the first two terms of a proportion be multiplied by
m, and the last two terms by w, the resulting products will
be in proportion.
Find the value of a: in the following proportions :
1. af 1: a; + 6 = 0? + 17: ajh 19.
2. 3a + 3:3aj4 = 6aj + l:5aja
3. x\a:x + b = x + c:x\'d.
4. mx + a : gaj + 6 = mx { c:qx + d.
6. (aj+7)(aj4) : (a;+3)(ajl) = (a;+l)(a;4) : (a?+l)(a;5).
Problbms.
1. Find a fourth proportional to 4, 6, 12.
2. Find a fourth proportional to ^, ^, ^.
3. Find a mean proportional to 4 and 9; 4 and 16.
4. Find a third proportional to 9 and 12; 7 and 14.
6. Find a mean proportional to  and ^; f and 30.
6. Find a third proportional to a^ and 2<ibi ocy and 3 ajy*.
7. Divide the number 20 into two parts such that the ratio
of their squares will be as 9 to 4.
8. Divide the number a into two parts such that the ratio
of their squares will be as m^ to n*.
§538.] VARIATION. 287
9. Divide the number 32 into two parts such that the
quotient of the greater divided by the less will be to that of
the less divided by the greater in the ratio of 25 to 9.
10. Find two numbers such that the sum of their squares
will be to the difference of their squares as 17 to 8, and the
difference of their squares to the difference of their cubes as
8 to 49.
11. Divide $ 121 among A, B, and C, so that A's share will
be to B's as 4 to 5, and B's to C's as 9 to 8.
12. The area of a rectangular field is 3 acres, and its length
to its breadth as 6 to 5. Find the dimensions.
13. The number of dollars A has is to the number B has as
9 to 6. By obtaining one half of A's money, B will have $ 2
more than A had at first. How much money has each ?
14. In a squarehewn block of stone containing 5 cubic
feet, the length is to the breadth as 9 to 5, and the breadth to
the thickness as 5 to 3. Find the dimensions of the stone.
15. A man's age is to that of his wife as 9 to 8. Ten years
ago their ages were as 13 to 11. What are their ages ?
16. Of two houses, one cost f 1000 more than the other, and
the ratio of their prices was as 3 to 2. Find the cost of each.
VARIATION.
537. When two quantities are so related that one increases
or diminishes in the same ratio as the other, the first is said to
vary as the second.
Thus, if a train of cars run a miles per hour, the distance
run in 2 hours will be 2 a miles, in 3 hours 3 a miles, and so
on ; that is, the distance will vary as the time varies.
538. This variation of two numbers is denoted by the
sign oc written between them, called the sign of variation, and
read " varies as." Thus, a oc 6 is read " a varies as 6."
288 ALGEBRA. [§ 639.
539. A number which in jiny particular problem cjianges its
value is called a yariable, and a number that has a fixed value
is called a constant. Thus, in the example given above, the
number of miles per hour (a) is a constant, while the numbers
denoting time and distance are both variables.
540. When two variables are so related that if one be given
the other can be found, one is said to be a function of the
other. Thus, for example, if a steamer sails at a given speed,
the distance sailed in a certain time will depend on the time,
and, if the time be given, the distance can be found; and
hence the distance is a function of the time.
541 . When one number varies as another, their correspond
ing values have a constant ratio.
Thus, if tti, Og, «3, etc., denote the increase of one variable,
and 6i, 62? ^s? ^tc, the corresponding increase of the other, so
that
^ = i, (l);and 5^=^«, (2)
then, from (1), ?Li = ^f? . ^nd from (2), ^ = ?i. (§ 525.)
bi 62 ^2 ^8
Hence i = ? = ^ . . . = a constant ratio.
h bi bs
Hence, if aocft, their common ratio is found by dividing
a by bf or a2 by 62, or Og by b^ and so on. If m denotes this
common ratio, 7 = m, and a = bm,
b
542. There are many ways in which two variables may be
related. Four of the more important cases are here presented.
543. I. When two numbers so vary that their corresponding
values have a constant ratio, they are said to vary directly.
Thus, if a Qc & directly, ^ = wi, their constant ratio, and
a = bm. Hence, if, in  = m or a= bm, a = 12 and 6 = 4,
m = 3 and a = 3b.
§ 545.] VARIATION. 289
1. A workman earns $40 a month. Upw much will he
earn in 5 months? In 12 months?
Which number in this problem is the constant ? Give the
two variables. Which is the function of the other ?
2. The base of a rectangle is 10 inches. What will be its
area if the altitude be 1 inch ? 3 inches ? 8 inches ?
Which number is the constant? Give the two variables.
Which is the function of the other?
544. II. When one number varies as the reciprocal of
another, the numbers are said to vary inversely. Thus, when a
varies inversely as 6, aoc  ; and then a = ~ x m, and ab = m.
For example, the time required to do a given work varies
inversely as the number of workmen employed. If 2 men can
do the work in 6 days, 4 men can do it in 3 days ; that is,
twice as many men will do it in one half of the time.
3. Three men can dig a ditch in 12 days. How long will it
take 6 men to dig it ? 9 men ? 12 men ?
Which number in this problem is the constant ? What num
bers are the variables ? Which is the function of the other ?
4. If a quantity of oats will feed 4 horses 9 days, how many
horses will it feed 3 days ?
545. In some cases one quantity varies inversely as the
square of another. Thus, aoc— •
For example, the illumination of a candle decreases inversely
as the square of the distance from it. If it gives a certain
illumination at a distance of 1 foot, the illumination at 2 feet
will be only \ as much ; at 3 feet, only ^ as much ; and so on.
6. If the illumination of a gas jet at 26 feet is a?, what will
it be at 60 feet ? At 100 feet ?
white's alo. — 19
290 ALGEBRA. [§ 546.
6. If the attraction of a magnet for a piece of iron at the
distance of ^ of an inch is x, what will be the attraction at ^^
of an inch ? ^ of an inch ? ^ of an inch ?
546. III. When one number varies as the product of two
other numbers, it is said to vary as the two others jointly.
Thus, if accbc, a varies jointly as 6 x c.
For example, the area of a rectangle varies as the product of
its base and altitude.
7. If the area of a rectangle with a given base and altitude
is X, what will be its area when the base and altitude are each
doubled ? When the base is trebled and the altitude doubled ?
The pressure of gas varies directly as its density, and also as
its temperature ; and hence, if the pressure of gas at a given
density and temperature is represented by x, its pressure when
its density is doubled, and its temperature increased one half,
will be expressed hyxx2x^ = Sx. Hence,
8. If the pressure of gas at a given density and temperature
is 15 pounds to the square inch, what will be its pressure if the
density be trebled, and the temperature reduced one half ?
547. IV. A number is said to vary directly as a second num
her, and inversely as a third, when it varies as the product of
the second and the reciprocal of the third.
Thus, a varies directly as 6, and inversely as c, when
a oc 6 X  ; that is, when a : 6 x  is constant,
c c
When a number varies as the quotient of two nambers, it varies
directly as the dividend, and inversely as the divisor. Thus, if ax, a
varies directly as &, and inversely as c. ^
9. The volume of gas varies as the absolute temperature, and
inversely as the pressure. If the volume is represented by x
when the pressure is 15 and the temperature 300, what will be
the volume when the pressure is 20 and the temperature 350 ?
§ 550.] VARIATION. 291
648. In all four of the cases of variation presented above,
the constant can be determined when any one set of corre
sponding values is given.
Thus (1), if aQC&, = w; (2), if aoc, ab = m; (3), if accbc,
a J /*\ 'i? r 1 h dc
— = m : and (4), if a oc o x , a =  x m. .. — = m.
be c c b
549. If a depends only on b and c, and aoc 6 when c is
constant, and aocc when b is constant, then, when both b and c
vary, a oc be.
Let a,b,c; a', b\ c, a", b', c', — be three sets of correspond
ing values.
Then, since c is in the first and second, —  =  • (1)
a' b'
also, since 6' is in the second and third,  = • (2)
a" c'
Multiplying (1) by (2), 5x^ = ^.5
^ a be . a a^^
whence — = — • .*. — = — •
a" 6'c' be b'e'
Hence a varies as be.
For example, the area of a rectangle varies as its altitude
when its base is constant, and as its base when its altitude is
constant, and as the product of its base and altitude when both
vary.
650. The simplest method of solving problems in variation
is to convert the variations into equations.
For example, if a oc 6 and 6 oc c, show that a oc c.
By § 543 we have a = bm (1), and b = cn (2), m and n being
constant ratios.
Multiplying (1) by (2), ab = bcmn ;
whence a = cmn, .*. aocc.
292 ALGEBRA. [§ 550.
10. If XQcy, and x = 5 when y = S, find x when y = 9.
11. If a<x h, and & is 9 when a is 6, what is h when a is 9 ?
12. a oc 6 and 6 x c : show that ococbK
13. a oc  and bcci show that a oc c
6 c
14. xcKyz: ii x = 2 when 2/ = 4 and 2 = 3, what will a equal
when y — 2 and z = 9?
15. If the area of a rectangle is x when its base is a and
its altitude by what will be its area when its base is 3 a and its
altitude 1 6 ?
16. The volume of a sphere varies as the cube of its diameter.
If the volume of a sphere 2 inches in diameter is 4.188 cu. in.,
what is the volume of a sphere 5 inches in diameter ?
17. The area of a sphere varies as the square of its diameter,
and the surface of a sphere 5 inches in diameter is 78.54 sq. in.
What is the surface of a sphere 10 inches in diameter ?
18. If the volume of a sphere varies as the cube of its diam
eter, how many spheres 3 inches in diameter equal a sphere 12
inches in diameter ?
19. The area of a circle varies as the square of its diameter.
How many circles 4 inches in diameter equal one 20 inches in
diameter ?
20. The velocity of a falling body varies as the time during
which it has fallen from rest. If the velocity of a falling ball
at the end of 2 seconds is 64 feet, what will be its velocity at
the end of 6 seconds ?
21. The distance a body falls from rest varies as the square
of the time it falls. ' If a ball falls 144 feet in 3 seconds, how
far will it fall in 12 seconds ?
22. The quantity of water that flows through a circular pipe
varies as the square of the pipe's diameter. If 10 gallons a
minute flow through an inch pipe, how many gallons per
minute will flow through a 4inch pipe ?
4 56«i.j PKOGRESSIONS. 293
CHAPTER XIX.
PROGRESSIONS.
551. A series is a succession of numbers formed according
to some fixed law, called the law of the series. The successive
numbers are called the terms. of the series.
552. A series that consists of a limited number of terms is
called a finite series, and one that consists of an unlimited
number of terms is called an infinite series. If a finite series be
considered as continued indefinitely in either or both direc
tions, it becomes an infinite series.
ARITHMETICAL PROGRESSION.
553. An arithmetical progression is a series in which each
term is obtained by adding a constant number to the preced
ing term. The constant number added is called the common
diiference. An arithmetical progression is denoted by A. P.
554. If the common difference is positive, the series is said
to be incjreasing. Thus the series 3, 5, 7, 9, 11, ••• is an increasing
arithmetical progression, in which the common difference is 2.
555. If the common difference is negative, the series is said
to be decreasing. Thus the series 12, 9, G, 3, 0, — 3, — 6, •••, is
a decreasing arithmetical progression, in which the common
difference is — 3.
556. The first and last terms of a finite arithmetical series
are called the extremes; and the intermediate terms, the
294 ALGEBRA. [§557.
arithmetical means. Thus, in the series 2, 6, 10, 14, 18, with
live terms, 2 and 18 are the extremes ; and 6, 10, and 14, the
arithmetical means.
557. If a denotes the first term of an arithmetical progres
sion, and d the common difference', then a f d will denote the
second term, a\2d the third term, a\Sd the fourth term ;
and so on to the nth term, which will be denoted by a(w— l)d.
If the series consists of n terms, and the nth or last term be
denoted by Z, we have the equation
l=a{'(n l)d (A)
558. This formula contains four symbols, denoting as many
different numbers. If any three of these numbers are given,
the fourth may be found by suhstitviing the given numbers for
their symbols in {A), and solving the resulting equations,
1. rind the thirteenth term of the A. P. 6, 10, 14, 18, ....
Here n = 13, a = 6, and <f = 10  6 = 4.
Substituting 13 for n, 6 for a, and 4 for d, in (^), we obtain
Z = 6 + (13  1) 4 = 6 + 12 X 4 = 54.
Hence the thirteenth or last term is 54.
The common difference in a given arithmetical series may evidently
be found by subtracting any term from the term which next follows it.
2. In an A. P. of 30 terms, the last term is 8, and the com
mon difference is 2. Find the first term.
Substituting 30 for n, 8 for Z, and 2 for d, in (^), we obtain
8 = a + (30  1) X 2.
.. a = 8 58 =50.
3. Find the number of terms of an A. P. whose first and
last terms are 5 and 65 respectively, and common difference 3.
Substituting 5, 65, and 3 for a, I, and d respectively, in {A), we obtain
65 = 5 + (n  1) 3 ;
whence n — 1 = 20. .*. n = 21.
§ 558.] PROGRESSIONS. 295
4. Find the common difference when the first and last terms
of an A. P. of 64 terms are 9 and 16 respectively.
Substituting 64, 9, and 16 for n, a, and I respectively, in (^), we obtain
16 = 9+ (64l)d.
, 169 1
•*• " = — :n — = :;;•
63 9
5. Insert 10 arithmetical means between 2 and 35.
Tlie first step is to find the common difference, and hence the problem
is similar to the above. Since there are 10 mean terms, the whole number
of terms is 12.
Substituting 12, 2, and 35 for n, a, and I respectively, in (^), we obtain
35 = 2 + (12l)d.
.. d = }f = 3.
Hence the required series is 2, 6, 8, 11, 14, 17, 20, 23, 26, 29, 32. 35.
6. The sixth term of an A. P. is 2, and the thirteenth term
is 23. What is the first term ?
Make the given sixth term the first term of a series of 8 terms in vv^hich
a = 2, 2 = 23, and n = 8.
Substituting 2, 23, and 8 for a, I, and n respectively, in {A), we obtain
23 = 2+ (8l)d = 2 + 7(Z.
.'. d = S.
Now consider the sixth term (2) the last term of a series of 6 terms,
and we have
2 = a + 5 X 3.
.*. a = — 13, the first term of the series.
7. Find the fifteenth term of the A. P. 17, 14, 11, ....
8. Find the sixteenth term of the A. P. — 12, — 7, — 2, «...
9. Find the fifteenth term of the A. P. 7, 6, 6^, ....
10. Given w = 15, d = 4, 1 = 70: find a.
11. Given n = 50, c? = f , ? = 5 : find a.
12. Given n = 105, d = .07, I =  2.36 : find a.
296 ALGEBRA. [§ 559.
13. Given cl = 3, a = 12, / = 72: find n.
14. Given d =  J, a = 5J, ? = 101: find*.
15. Given d = .5871, a = .29, i = 59 : find n.
16. Given 11 = 22, tt = 4, /=130: find d.
17. Given n = 58, a = .33, / = 75: find d.
18. Insert 7 arithmetical means between 5 and 29.
19. Insert 8 arithmetical means between 1 and — 5.
20. Insert 9 arithmetical means between — 6 and 0.
559. A general formula for finding the sum of n terms of an
arithmetical progression may be obtained as follows:
Let s denote the sum of the terms, and I the last term ; then
« = a + (a + d)h(a + 2d) + (a + 3d).. f^ (1)
also s = ;f (Zd)4(i2d)f (Z3d)... +0, (2)
the sura in (2) being written in reverse order.
Adding (2) to (1), term to term, we obtain
25= (a40 + («+0 + («+0+(«+0H h(aH0 ton terms.
Hence 2s = n(a^l),
and « = ^(a + r). (B)
560. This formula also contains four symbols, and may be
employed to obtain solutions for as many classes of problems.
21. Find the sum %i 15 terms of an A. P. that has for its
extremes 7 and 45.
Here n = 15, a = 7, Z = 45; and substituting these values for n, a,
and I respectively, in (i?), we obtain
« = \ (7 + 45) = 15 X 26 = 390.
it
§ 562.] PROGRESSIONS. 297
22. The sum of 40 terms of an A. P. is — 660, and the last
term is — 36. Find the first term.
Substituting 40, — 560, and — 36, for n^ s, and {, in (jB), we obtain
660 = — (a 36),
whence ^^720560^3^
20
The last term of an A. P. is found in like manner when s, a, and n are
given.
23. The sum of the terms of a series is 272, and the ex
tremes are — 13 and 45. Find the number of terms.
Substituting — 13, 46, and 272, for a, I, and s respectively, in (5), we
obtain
272 = I (13 + 46),
1 272 ,^
whence n = — = 17.
16
24. Given a = 11, 1 = 25, n = 14: finds.
25. Given a = 13, Z = 105, n = 63: finds.
26. Given a = — 3.7, I = — 1.3, w = 42 : find s.
27. Given s = 598, Z = 39, w = 26 : find a.
28. Given s = 1295, Z = 80, n = 37 : find a.
29. Given s = 1184, a = 11, n = 32 : find Z.
30. Given s = 116.375, a =  3.5, w = 19 : find Z.
31. Given s = 336, a = 3, 1 = 25: find w.
32. Given s = 30.225, a = .75, Z = 1.2 : find w.
33. Given s = 147 V5, a = 3 V5, Z = 18V5: find w.
561. The fundamental formulas (A) and (B) contain, in all,
five symbols; and such is the relation between the numbers
represented, that, if any three of them are given, the other two
can be found.
562. Of the twenty classes of problems thus arising, eight
can be solved by a single formula, as above exemplified; but
the remaining twelve classes require both formulas, since they
298 ALGEBRA. [§ 562.
each involve the numbers denoted by d and », given or required,
and d occurs only in {A)y and 8 only in (B).
34. Find the sum of 30 terms of the series 7, 10, 13, •••.
Here n = 30, a = 7, and d = 3, and s is required.
Substituting their values for n, a, and d, in {A) and (£), we obtain
I = 7 + (30  1) X 3 = 04, (1)
8 = V (7 4 = 105 + 151. (2)
Substituting in (2) the value of Z in (1), we have
s = 105 + 15 X 94 = 1515.
f
35. How many successive terms of the series 11, 9, 7,
must be taken, that their sum may be 27 ?
Here a = ll,d = — 2, and s = 27, and n is required.
Substituting their values for a, d, and s, in {A) and (B), we obtain
1 = 11 + (»l)xC2)=132n, (1)
27 = (ll + 0 .. 54 = lln+Z». (2)
Substituting in (2) the value of I found from (1), simplifying, etc., we
obtain n212n = 27;
solving the equation, n = 9 or 3.
For n = 9, the A. P. is 11, 9, 7, 5, 3, 1,1, — 3, — 5, the sum being
27.
For n = 3, the A. P. is 11, 9, 7, the sum being 27.
To find the value of I, substitute both values of n in (1), obtaining, for
n = 9, Z = 13  2 X 9 =  5 ; for n = 3, Z = 13  2 x 3 = 7.
PROBLEMS.
Find the sum of
1. 30 terms of the A. P. 5, 7, 9, ....
2. 16 terms of the A. P. 9, 6, 1, ....
3. 36 terms of the A. P. 1^, 2, 2^, ....
4. 15 terms of the A. P. — 2J, ~1, —1, ....
5. 19 terms of the A. P. 5.3, 6.1, 6.9, .... .
6. 100 terms of the A. P. 1, 2, 3, ....
7. n terms of the A. P, 1, 2, 3, •••,
§ 562,] PROGRESSIONS. 299
8. Find the sum of n terms of the A. P. 1, 3, 5, •••.
9. Find the nth term of each series in 7 and 8.
10. The extremes of an A. P. are 2 and 30, and the common
difference is 2. Find the sum of the series.
11. The first term of an A. P. is 10, the number of terms 10,
and the sum of the terms S5, Find the common difference.
12. The common difference of an A. P. is 2, the last term
23, and the number of terms 11. Find the sum of the series.
13. The common difference of an A. P. is 2, the number of
terms 12, and the sum of the terms 90. Find the last term.
14. The number of terms of an A. P. is 13, the common dif
ference — 7, and the sum of the terms 39. Find the first term.
15. The last term of an A. P. is 52, the common difference
5, and the sum of the series 297. Find the number of terms.
16. The fourth term of an A. P. is — 1, and the tenth term
is 3. What is the first term ? What is the sum of the series ?
17. A man paid $3000 in 6 installments. The first was
$ 400, and the last $ 600, all the payments being in an
A. P. What was the amount of each intermediate payment ?
18. The cost of sinking a well was $ 45, f 1 being paid for
sinking the first yard of depth, $ 1.50 for the second, $ 2 for
the third, and so on. What was the depth of the well ?
19. A man began saving by putting by 1 cent on New Year's
Day, 2 cents on the next day, 3 cents on the next, and so on.
In how many days would he have put by f 98.70 ?
20. A load of 100 fence posts was laid down at the corner
of a field. If, leaving 1 post at the corner, a man should
carry the rest one by one, to lay them down in a straight line
at intervals of 3 yards, how far would he walk to accomplish
his task ?
800 ALGEBRA. £§ 563.
GEOMETRICAL PROGRESSION.
563. A geometrical pnogression is a series in which each term
bears a constant ratio to the preceding one. A geometrical
progression is denoted by G. P.
564. This constant or common ratio is called the ratio of the
aeries ; and the series is increasing or decreasing according as
its ratio is numerically greater or less than unity.
Thus, the series 2, 6, 18, •••, is an increasing geometrical
progression in which the ratio is 3; and 64, 16, 4, •••, is a
decreasing progression in which the ratio is \.
565. If the ratio be positive, all the terms will be positive;
but if the ratio be negative, the terms will be alternately posi
tive and negative, as in the series 5, — 10, + 20, — 40, • • •, in
which the ratio is — 2.
A geometrical progression is a continued proportion, as defined in § 520.
566. When a geometrical progression consists of a definite
number of terms, the first and last are called eztrQmes ; and the
intermediate terms are the geometrical means.
A geometrical scries may be considered as extending indefinitely in
either direction, and it is then called an infinite geometrical series,
567. If a denotes the first term of a G. P., and r the ratio,
then ar will denote the second term, aii^ the third term, ai^
the fourth term; and so on to the nth term, which will be
denoted by ar^'^, it being the (n — l)th term after the first.
Hence the last term (/) in a series of n terms will be desig
nated by the formula
I = ar^\ {A)
568. This formula, like the corresponding one in arith
metical progression, contains four symbols, denoting as many
§ 568.] PROGRESSIONS. 801
numbers ; and, if any three of these numbers are given, the
fourth can be found hy substituting the given numbers for their
symbols in (A), and solving the resulting equation.
1. Find the ninth term of the G. P. 243, 81, 27, ....
Here a = 243, n = 9, and r = ^\^ = J.
Substituting their values for a, n, r, in (^), we obtain
i = 243 X f i V= 243 X = — •
\SJ 243 X 27 27
It is evident that the ratio in a given G. P. is found by dividing any
given term by the one preceding it.
2. Find the first term of the G. P. whose eleventh term is
3072, and whose ratio is — 2.
Substituting their values for n, ?, r, in (A), we obtain
3072 = a X ( 2)10 _ « x 1024 ;
whence a = 3072 f 1024 = 3.
3. Find the ratio of the G. P. whose first and eighth terms
are 1715 and yttt respectively.
Substituting their values for a, I, w, in (^), we obtain
^^^=1715r7;
6 1
whence r =
2401 X 1715 2401 x 343
=^^
1
2401 X 343 7
4. Insert 4 geometrical means between — ^ and 144.
The first step is to find the ratio, as in the preceding problem. The
number of terms is 4 + 2, or 6.
Substituting their values for a, 2, w, in (A), we obtain
144 = ^Vx^,
whence . r = — 6.
Hence the required means are
A^ ih W»  4F; or J, }, 4, 24.
5. Find the eighth term of the G. P. 1, 2, 4, •••.
6. Find the tenth term of the G. P. 28, 7, J, ....
S02 ALGEBRA. [§ 56a
7. Find the seventh term of the G. P. 1, — ^, t^, •••.
8. Given I = 1458, w = 6, r = 3 : find a.
9. Given I = 960, n = 7, r = — 2: find a.
10. Given Z = — 567, n = 5, a = — 7 : find r.
11. Given Z = f, n = 7, a = 2: find r.
12. Insert 3 geometrical means between 2 and ff.
13. Insert 7 geometrical means between ^ and 9.
14. Insert 5 geometrical means between ^ and jTTg^.
15. Show that the geometrical mean between two numbers
is the square root of their product.
669. When a, I, and r are given in order to find n, the pro
cess leads to an exponential equation; that is, an equation in
which the unknown number occurs as an eocponent. The solu
tion of this form of an equation can usually be effected only
by the use of logarithms, as shown in § 613.
570. A general formula for finding the sum of n terms of a
G. P. may be obtained as follows :
Let s denote the sum of the terms ;
then s = a h ar h ar* 4 ar^  ••• + ar'^'K (1)
Multiplying by r, rs = ar\ ar^  ar^ + '•• f aY^~^ + ai^. (2)
Subtracting (1) from (2),
r8 — 8 = ait^ — a, or s (r — 1) = a (r* — 1) ;
«^ 1
whence s = a x (B)
r — 1
671. This formula also contains four symbols, and may be
employed for the solution of as many classes of problems ; but
in two cases it may lead to equations of a higher degree than
the second.
§ 573.] PROGRESSIONS. 303
16. Find the sum of 8 terms of the G. P. 2, 6, 18 ..
Here a = 2, n = 8, r = 3.
Substituting their values for a, n, r, in (B), we obtain
s = 2 X ^^^^ = 38  1 = 6560.
31
17. The sum of 10 terms of a G. P. is — 1705, and the ratio
is — 2. Find the first term.
Substituting their values for n, r, 8,iD. (B), we obtain
1706 = ax^^^^ — ^ = ax 341;
— ^ — 1
whence a = — = 6.
341
18. Find the sum of 6 terms of a G. P. the first and last
terms of which are 3 and ^ respectively.
Substituting their values for n, a, I, in (^), we obtain
r = i.
Substituting their values for a, n, r, in (B), we obtain
« = 3 X W = 5l
672. When r < 1 numerically, the formula (B) is more con
venient in the form
1 — r»
5 = a X
1 — r
When r = l, 8 = a(l + r + r^ + r8f... + r»^), and hence s = an.
Infinite Series.
573. If r < 1, the value of r" decreases as n increases ; and
hence, as n is indefinitely increased, r** is indefinitely dimin
ished, and thus approaches zero for its limit Therefore the
above formula becomes a
—^ — being the limit towards which s approaches when r < 1,
1 — r
and both a and r are positive, and n is indefinitely increased.
This limit is properly called the limit of the sum of a
804 ALGEBRA. [§ 573.
converging geometrical series as n approaches infinity, but for
convenience it is usually called the limit of the series to itifinity.
An infinite number is denoted by the symbol oo.
For converging and diverging series, see Chapter XXI., p. 336, note.
19. Find the limit of the sum of 1 + ^ f ^ ••• to infinity.
Here a = 1, r = J ; and hence, by (B'), 8= =2.
1 — i
This may be illustrated by taking a slip of paper, say 2 inches long,
dividing it into two equal parts, taking away one of these parts, and
bisecting tlie other ; and so on indefinitely. It is evident that the sum of
the parts taken away can never exceed 2 inches.
20. Find the sum of 10 terms of the G. P. 2  6 + 18 .
21. Find the sum of 7 terms of the G. P. 75 + 15 h 3 + .%
22. Find the sum of 8 terms of the G. P. 1 —  + f ..
23. Given w= 8, r= 2, s = 765: find a.
24. Given n= 7, r=— 3, « = 547 : find a.
25. Given n = 10, r = !> * "^ ^^ff • ^^^ ^•
26. Given n = 7, r = .5, s = 1943.1 : find a.
Find the limit of the sum, to infinity, of
27. 1 + ^ + ^+.... 30. 3 + ff^+.
28. 1+1 + 1+.... 3,^ l+l + i. ....
29. f +  + A+. ^ ^
32. Find the value of the recurring decimal .53434 + •...
The repeating part, .03434 +, is an infinite geometrical series, the first
term being ^Ujj^ ^*^® second rjjjjjy^, and so on.
Substituting ^ooir for a, and yj^ for r, in {B'), we obtain
Hence the value of the decimal is /^^ + ^^q = UJ.
A recurring decimal is also called a circulating or a repeating decimal.
The repeating part, called the repetend, is denoted by a dot placed over
the first and the last of its figures ; thus, .534.
§ 575.] PROGRESSIONS. 305
Find the value of
33. .24i. 35. .354. 37. 8.90l.
34. 6.53. 36. .45i24. 38. 54.321. (See also ^ 6SS.)
574. Of the twenty possible cases in geometrical progres
sion, eight can be solved by either (A) or (5), as shown above.
The twelve remaining cases — that is, all those cases in which
both the numbers denoted by I and s are involved — require the
use of both formulas.
39. The extreme terms of a G. P. are 3 and 192, and the
ratio 2. Find the sum of the terms.
Substituting their values for a, Z, and r, in (A) and (5), we obtain
192 = 3 X 2»i. .. 2"i = 64, or 2« = 128. (1)
s = 3 X ^*~ ^ ' .'. substituting 128 for 2», s = 381. (2)
40. Find the seventh term of the G. P. whose ratio is f , and
the sum of the series 42.
Substituting their values for n, r, «, in (4) and (B), we obtain
i = ax(i)6 = axW» (1)
2059 W/1 2059 . ,^
__=axI^ = ax^. .. a = t. (2)
Substituting J for a in (1),
575. When the number denoted by n is required, the process
leads to an exponential equation (§ 569) ; and in some cases, if
n > 2, to equations of higher degree than the second. The
foregoing examples may sufficiently indicate how to proceed in
soluble cases. Sometimes the value of n in an exponential
equation is readily found by inspection, as in 2" = 128, occur
ring in the solution of Example 39 above.
WHITR*S ALO. 20
306 ALGEBRA. [§ 675.
PROBLEMS.
1. Given a = 4, 1 = 2916, n = 7: find s and r.
2. Given a = 5, 1 = 320, n = 7 : find s and r.
3. Given 7' = 4, Z = 1024, 7i = 9: find s and a.
4. Given r = 2, Z = 20480, n = 14; find a and s.
5. Given r = 3, s = 2391484, a = ^: find Z.
6. Given i = 65536, s = 74898^, a = ^: find r and w.
7. What is the sum of the series 2, ^, f, •••, to infinity ?
11 1
8. What is the sum of 1, — — , — , — ^> •••> to infinity ?
9. The extremes of a G. P. are 2 and ^^, and the ratio
is ^. What is the sum of the series ?
10. The sum of the third and fourth terms of a G. P. is 18,
and the difference of the third and fifth terms is — 36. Find
the series.
11. A population increases yearly in geometrical progression,
and in 4 years is raised from 10,000 to 14,641. Find the ratio
of annual increase.
12. There are 3 numbers in geometrical progression whose
sum is 62 ; and the sum of the first and second is to the sum
of the second and third as 1 to 5. Find the numbers.
13. The difference between the first and second of 4 num
bers in geometrical progression is 15, and the difference
between the third and fourth is 540. Find the numbers.
14. A person wishes to send f 9950, besides as much more
as will cover the express charges for the whole sum at the rate
of ^ %. How much should he send in all ?
15. Some grains of wheat found in the crop of a wild fowl
being planted, one germinated and produced 50 sound grains.
These again being sown produced a crop of 2500 grains. Of
how many grains would the crop of the sixth year consist, sup
posing the grain to increase every year at the same rate ?
§ 578.] PROGRESSIONS. 807
16. An elastic ball falls from a sufficient height to rebound
30 feet, and at each successive rebound rises ^ of the distance
of the previous one. How many feet will the ball pass over in
5 rebounds ? How many feet before it comes to rest ?
17. There are 4 numbers in geometrical progression, and
the first is 21 less than the fourth, and the difference of the
extremes divided by the difference of the means is 3^. Find
the numbers.
HARMONIC PROGRESSION.
676. Three numbers are said to be in harmonic proportion
when the first is to the third as the difference of the first and
second is to the difference of the second and third.
Thus, if a, h, c, are in harmonic proportion,
a\c = a — b\h — c, or a\c=h — a\c — h.
yt*t. An harmonic progression is a series in which the first
of any three consecutive terms is to the third as the difference
of the first and second is to the difference of the second and
third. An harmonic progression is denoted by H. P.
Thus, if a, h, c, d, e, are in harmonic progression,
a:c=:a—b:h—c\ b:d=b—c:c—d; and c: e=c— d: d— e.
578. If a, b, c, are in harmonic progression, we have, by
definition,
a:c = a — 6:6 — c;
whence c(a — b) = a(b — c).
a — b b — c
Dividing by abc,
whence
ab be *
1111
b a c b
Hence, if given numbers are in harmonic progression^ their
reciprocals are in arithmetical progression.
308 ALGEBRA. [§ 579.
579. Problems relating to harmonic series are usually best
solved by writing the reciprocals of the terms as an arithmetical
progression, and then solving the resulting equations.
1. The second term of an H. P. is 2, and the fourth term 6.
Pind the series.
The second aud fourth terms of the corresponding A. P. are } and (.
Let a be the first term, and d the common difference. Then
o + d = i,
whence o = J, and d = — \.
The A. P. is }, J, J, i, etc.;
hence the H. P. is f , 2, 3, 6, etc.
2. Insert 3 harmonic means between 3 and 16.
The first and fifth terms of the corresponding A. P. are ^ and ^ ; and
hence d = (^^  J) ^ 4 =  ^J^.
Thus the A. P. is J, A, ft» A» A
Hence the 3 harmonic means are ^^, 5, ^.
3. The first term of an H. P. is 2, and the third term is 6.
What is the second term ?
4. The second term of an H. P. is &, and the third term is c.
What is the first term ?
6. The first term of an H. P. is 2, and the fourth term is 6.
Find the mean terms.
6. Insert 5 harmonic means between \ and j^.
7. Insert 4 harmonic means between f and f.
8. Insert 4 harmonic means between \ and j^.
9. The first term of an H. P. is 1, and the third term ^.
Pind the sixth term.
10. , , , are in arithmetical progression. Show that
a h c
a — b :b — c = a:a
§ 582.] LOGARITHMS. 809
CHAPTER XX.
LOGARITHMS.
580. The logarithm of a number is the exponent of the
power to which a fixed number, called the base, must be raised
in order to produce the given number.
Thus, since 3* = 81, 4 is the logarithm of 81 to the base 3;
since 8* = 512, 3 is the logarithm of 512 to the base 8 ; and,
generally, if a* = m, x is the logarithm of m to the base a,
581. The base of the system of logarithms in common use is
10, the basis of the decimal notation ; and the system is called
the common system.
682. Since 10^ = 1, 10  ^ = 1 = .1,
' 10 '
10^ = 10, 102 = L=.oi,
10^ =100, 10 3 = L = .001,
> 103 ^
10« = 1000, 10* = i= .0001,
' 10* '
10* = 10000, and so on,
the numbers 0, 1, 2, 3, •••, are the logarithms of the successive
positive integral powers of 10 ; and — 1, — 2, — 3, — 4, •••, are
the logarithms of the successive negative integral powers of 10.
Thus (log being an abbreviation for logarithm),
log 1 = 0, log.l =1,
log 10 = 1, log .01 =2,
log 100 = 2, log .001 =  3, and so on.
810 ALGEBRA. [§ 583.
583. The logarithms of numbers between the integral
powers of 10 are evidently fractional. Thus, the logarithms
of the numbers between
100 and 1000 are 2 h a decimal,
10 and 100 are 1 h a decimal,
1 and 10 are f a decimal,
1 and .1 are — 1 + a decimal,
.1 and .01 are — 2 h a decimal,
.01 and .001 are — 3 f a decimal,
and so on.
Thus, with — above negative characteristics, the logarithm of
5 is 0.69897, .5 is 1.69897,
25 is 1.39794, .025 is 2.39794,
225 is 2.35218, .00225 is 3.35218.
684. It is evident from the above that a logarithm consists
of two parts : (1) a positive or negative integral number, called
the characteristic ; and (2) a positive fractional part, called the
mantissa. Thus, in the logarithm 2.78176 (log 605), the 2 is
the characteristic, and the decimal .78176 is the mantissa.
The characteristic is so called because it shows, as will be explained
later, the number of orders, integral or decimal, in the corresponding
number.
685. It is also evident that the characteristic of the log
arithm of a number greater than 1 is positive^ and that the
characteristic of the logarithm of a number less than 1 and
greater than is negative. The mantissas of all logarithms
are positive.
586. There are three ways of writing a logarithm when its
characteristic is negative, as follows :
(1) 2.41162; (2) .411622; (3) 8.4116210.
12.33244 ; .33244  12 ; 8.33244  20.
§ 590.] LOGARITHMS. 311
The first method is simple, and is used herein. The sign —
is written over the characteristic to show that it alone is nega
tive, the mantissa being always positive.
587. The number that corresponds to a given logarithm is
called the antilogarithm, which is abbreviated as antilog.
Thus, 605 is the antilog of the logarithm 2.78176.
588. The logarithms of a series of numbers arranged in
tabular form is called a table of logarithms (§ 611).
The logarithms of numbers, as now found arranged in tables, were
originally calculated by very laborious methods, consisting essentially in
repeated extractions of the square root. These methods now possess only
an historic interest, since methods have been devised by which logarithms
can be calculated with great facility. These methods, however, depend
upon principles the proof of which belongs to a more advanced treatise
on algebra.
PRINCIPLES.
589. The use of logarithms as a means of facilitating certain
numerical computations involves principles which we now pro
ceed to establish; and since these principles are the same,
whatever may be the base of the system, let us denote the
base by the general symbol a.
690. Let m and n denote any two numbers, and x and y
their respective logarithms to the base a; then, by definition,
a' = m. (1)
a' = n, (2)
Multiplying together (1) and (2) member by member, we have
a*+y = mn ;
whence x + y = log mn, by definition.
Hence the logarithm of a product is equal to the sum of the
logarithms of the factors.
It follows that the logarithm of a composite number is the
sum of the logarithms of its factors.
312 ALGEBRA. [§ 591.
691. Dividing (1) by (2) member by member, we have
a"" = — ;
n
whence x — y = log—
n
Hence the logarithm of a quotient is equal to the difference of
the logarithms of dividend and divisor.
It follows that the logarithm of a common fraction is the
logarithm of its numerator minus the logarithm of its denomi
nator.
The sums and differences of numbers cannot be found by logarithms.
592. Raising both members of (1) to the power denoted by
J), we have
a*" = m';
whence px = logm^.
Hence the logarithm of a power is equal to the logarithm of
the bajie of that power multiplied by the exponent of the power.
593. Extracting the rth root of both members of (1), we have
a'^ = S/m ;
whence  = log S/m.
r
Hence the logarithm of a root is equal to the logarithm of the
power divided by the index of the root,
594. If in (1) we make m = 1, the corresponding value of x
will be zero, since a° = 1, whatever a may be (§ 123).
Hence, in aU systems of logarithms^ logl = 0,
595. If, again, in (1) we make x = l, then, since a^ = a, or
log a = 1, it follows that in all systems of logarithms the log
arithm of the base is unity.
§ 599.] LOGARITHMS. 813
596. The application of the foregoing principles to numerical
computations by logarithms involves (1) the obtaining of the
logarithms of numbers from the tables, and (2), when a loga
rithm is given, the finding of the corresponding number or
antilog. These processes will now be explained.
ARRANGEMENT AND USE OF TABLES.
597. The simplest method of presenting a table of logarithms
would be to arrange the numbers in their natural order in ver
tical columns, and place opposite each number its logarithm;
but tables thus arranged would be altogether too voluminous
for use. Various expedients, therefore, are employed in order
to save space in tables, and also time in their use. The great
advantage, indeed, of adopting 10 as the base of the common
system arises from the fact that tables of logarithms to that
base can be presented in a very compact form.
698. It has already been shown that the logarithm of any
number between two integral powers of 10 lies between the
integers which are the logarithms of those powers (§ 583).
Thus, the logarithm of the number 3265, which is between 10^
and 10*, must lie between 3 and 4, the exponents of the given
powers of 10. This logarithm is, in fact, 3.51388.
It is thus seen that the positive characteristic of a logarithm
is always one less than the number of integral orders in its
antilog.
699. The mantissas of a series of logarithms increase as
their antilogs increase from one power of 10 to another. This
increase of mantissas is shown below for the antilogs from 1
to 10^ and from 10^ to 10^, only three figures of each mantissa
being given.
..rl 2 3'4 5 6 7 8 9 10
Antilog I ^^ 2^ ^^ ^^ ^^ g^ ^^ g^ ^^ ^^^
Mantissa .000 .301 .477 .602 .698 .778 .845 .903 .954 .000
814 ALGEBRA. £§ 60a
600. The numbers 326500, 32650, 3265, 326.5, 32.65, 3.265,
.3265, .03265, etc., expressed by the same sequence of figures,
but differing in the position of the decimal point, may all be
derived from any one of the set by multiplying or dividing by
some power of 10. Hence the logarithms of any two of these
numbers can differ from each other only by tlie logarithm of
some power of 10; that is, by some integer, positive or nega
tive, the mantissa remaining the same for all. Hence
Numbers expressed by the same sequence of significant figures
have the same mantissa in their logarithms.
Thus, log 326500 = log (3265 x lO^) = log 3265 f log 10*
= 3.51388 + 2 = 5.51388.
log .03265 = log (3265 ? l(f)_ = log 3265  log 10*
= 3.51388  5 = 2.51388.
601. Since the increase or decrease of the characteristic by
so many units manifestly corresponds to the shifting of the
decimal point so many places to the right or left in the anti
log, what has been shown above in regard to one series of
antilogs and their logarithms will evidently hold good in
regard to any other similarly related series.
602. It follows that the characteristic of the logarithm of any
number is equal to the number of places by which the lefthand
digit of tJiat number is distant from the units place; and that
it is positive when the digit lies to the left, iiegative when it
lies to the right, and zero when it is in the units place.
Thus, (1) log 3625 = 3.51388, (5) log .3625 = 1.51388,
(2) log 362.5 = 2.51388, (6) log .03625 = 2.51388,
(3) log 36.25 = 1.51388, (7) log .003625 = 3.51388,
(4) log 3.625 = 0.51388, and so on.
603. It may be adopted as a good rule, that a positive char
acteristic is numerically less by 1 than the number of figures
before the decimal point of the given number; and that a
§ 605.] LOGARITHMS. 815
negative characteristic is numerically greater by 1 than the
number of zeros after the decimal point and before the first
significant figure. Thus, the characteristic of log 73582 is 4 ;
the characteristic of log 4.965 is 0; and the characteristic of
log .00053 is 4.
As the characteristic can thus always be supplied by in
spection of the given number, it is accordingly omitted from
the tables, which might more appropriately be called tables
of mantissas than tables of logarithms.
Write down the characteristics of the logs of the following
numbers, placing the minus sign over negative characteristics :
1. 7; 70; 70000; .7; .00007. 3. 5360; .536; 5.36; .00536.
2. 23; 2.3; .23; 230; .023. 4. 43892; 43.892; 43892000.
604. The mantissas in the tables herein given (pp. 319321)
may be regarded as those of the logarithms of all numbers from
1 to 1000. The mantissa of log 365, for example, is found in
the line opposite 36, the first two figures of the antilog, and
in the column headed by 5, the third figure. The mantissa
thus found is .56229.
In the line opposite 36, in the columns headed by 0, 1, 2, ... 9, we find
, the mantissas .55630, .55751, .55871, .•• .56703, belong respectively to the
logs of 360, 361, 362, ... 369. In the line opposite 40, the mantissa in
the column headed is not only the mantissa of log 400, but also of
log 40, log 4, log .04, etc.
605. To find the logarithm of a number expressed by not
more than three figures.
To the proper characteristic annex the mantissa found opposite
the first twofi^gures in the column headed by the third.
Thus,
(1) log 300 = 2.47712; log 30 = 1.47712; log 3 = 0.47712 ;
(2) log .03 = 2.47712 ; log 57 = 1.75587 ; log 579= 2.76268 ;
(3) log 4.78 =0.67943; log 10.5=1.02119; log 2390 =3.37840.
Note. Always write down the characteristic before seeking the
mantissa.
316 ALGEBRA. [§ 606.
Find the logarithms of the following numbers:
6. 2, 3, 4, .5, .6, .007, 80, 900, 90000.
6. 23, 340, 4500, 5.3, .67, .072, .0085.
7. 121, 306, 272000, 35.4, 4.62, .537, .00643.
8. .876, 7.65, 65.4, 543, 4320, 32100.
606. The method of finding the logarithms of numbers ex
pressed by more than three figures is made evident by a single
illustration. Since the number 362.45, which is greater than
362, is less than 363, the difference between log 362 and log 363
is greater than the difference between log 362 and 362.45. Let
d denote the difference between log 362 and log 363; then
log 362.45 will be log 362 increased by less than d.
Assuming that the logarithmic difference for a fraction of a
unit will be the same fractional part of the difference for a
whole unit, — an assumption generally true as far as two places
of decimals, — we can find log 362.45 by multiplying d by .45,
and adding the product to log 362. But log 362, as found in
the tables, is 2.55871, and the difference between, log 362 and
log 363 is .00120, and hence log 362.45 is obtained as follows :
log 362 = 2.55871
d X .45 = .00120 X .45 = 54
whence log 362.45 = 2.55925
Since the mantissa for log 36245 is the same as that for
362.45 (§ 600), we may find log 36245 by considering 45 a
decimal, and proceeding in exactly the same way, except that
we make the characteristic 4 instead of 2. Hence,
607. To find the logarithm of a number expressed by more
than three figures,
To the logaritJim of the number expressed by the first three
figures, as found in tlie table, add the product of the corresponding
§ 608.] LOGARITHMS. 317
tabular difference multiplied by the number expressed by the
remaining figures regarded as a decimal.
In the tables of logarithms given below, the difference between each
mantissa and the next higher mantissa is printed over the first. These
differences are called tabular differences,
9. ¥ind log 4.7389.
log4.73 = 0.67486 d = 92
d X .89 = .00092 X .89 = 82 JB9
whence log 4. 7389 = 0.67568 828
736
d' = 81.88
For convenience the tabular difference may be regarded as an integral
number, and the product be formed as at the right above. We reject the
88, and since the first rejected figure is not less than 5, we consider the
81 as 82. Generally increase the fifth figure by 1 when the sixth figure of
a mantissa is more than 5, and drop the sixth figure when it is less than 6.
10. Find log .00124.
log .00124 = 3.09342
d X .6 = 349 X .6 209
whence log .00124 = 3.09551
Find the logarithms of the following numbers :
11. 96520; 8736; 764.5; 65.48; 5.321; .004512.
12. 35962; 2045.3; 156.78; 93.125; 8.6545.
13. .004567; .0056789; .06789; .78901.
14. .89012; 90.123; 123.45; 2345.6; 34567.
608. The antilog of a given logarithm is found by practi
cally reversing the above process. For example, the antilog of
4.81491 is obtained (1) by finding the mantissa in the tables,
and writing the corresponding antilog. The mantissa .81491
is found in line 65, column 3, and hence the antilog of said
mantissa is 653. But (2) since the characteristic of the given
logarithm is 4, there must be five (4 + 1) integral orders in the
818
ALGEBRA.
[§609.
antilog, and hence the antilog sought is 65300. If the given
characteristic had been 2, the antilog would have been 653;
and, if the characteristic had been 0, the antilog would have
been 6.53 ; and so on.
609. To obtain the antilog of 2.89387, we find in the tables
the next lower mantissa, which is .89376, in line 78, column 3.
We write 783 for the first three figures of the antilog. The
remaining figures are found as below :
Given mantissa, .89387
Next lower mantissa, .89376
Dividing by d or 55, 55)11.00(.20
Annexing the quotient .20 to 783, and pointing so as to have
three (2  1) integral orders, we obtain 783.2 as the required
antilog. If the characteristic had been 2, we should have
placed one cipher (— 2l) after the decimal point, thus obtain
ing .07832 as the antilog.
610. To find the antilog of a given logarithm,
I. If the given mantissa is found in the tables, write down
the figures of the corresponding antilog, and determine the posi
tion of the decimal point by the given characteristic,
II. If the exact given mantissa is not found in the tables, sub
tract from it the next loiver mantissa, and divide the difference by
the tabular difference of this lower mantissa. Annex the first two
figures of the quotient to the antilog of the lower mantissa, and
place the decimal point as determined by the given characteristic.
Find the antilogs of the following logarithms :
15. 0.30103
16. 1.47712
17. 1.07918
18. 2.13113
19. 3.64098
; 0.47712;
, 0.60206;
; 2.69897
; 3.77815;
; 1.36173
; 1.65321;
; 2.51388;
2.77633 ;
; 4.77541
; 2.82816;
0.84510; 0.95424.
3.90309; 2.95424.
1.81954; 1.92428.
1.86435; 1.96904.
0.91172; 4.72652.
611. TABLE OF LOGARITHMS: with Tabular Differekceb.
N
O
1
2
3
4
5
6
7
8
432
42S
424
419
416
412
408
404
400
396
10
00000
00432
(xmo
01284
01703
02119
02531
02938
03342
03743
393
389
386
382
379
376
372
369
366
368
11
04139
04532
04922
05308
05690
06070
06446
06819
07188
07555
360
857
355
351
849
346
843
344)
338
885
12
07918
08279
08636
08991
09342
09691
10037
10380
10721
11069
833
330
828
325
823
321
318
315
314
811
13
11394
11727
12057
12385
12710
13033
13354
13672
13988
14301
809
307
305
303
301
299
297
295
298
291
14
14613
14922
15229
15534
15836
16137
16435
16732
17026
17319
289
287
285
2S8
281
279
278
276
274
272
15
17609
17898
18184
18469
18752
19033
19312
19590
19866
20140
271
269
267
266
264
262
261
259
258
256
16
20412
20683
20952
21219
21484
21748
22011
22272
22631
22789
254
253
252
250
249
248
246
245
243
242
17
23045
23300
23553
23805
24055
24304
24551
24797
25042
25286
241
2:39
238
2:37
235
234
233
232
230
229
18
25627
25768
26007
26246
26482
26717
26951
27184
27416
27646
228
227
226
225
223
222
221
220
219
218
19
27876
28103
28330
28656
28780
29003
29226
29447
29667
29885
217
216
2 15
213
212
211
210
209
208
207
20
30103
30320
30535
30750
30963
31176
31387
31697
31806
32016
206
205
204
203
202
202
201
200
199
198
21
32222
32428
32634
32838
33041
33244
33446
33646
a3846
34044
197
196
195
194
193
193
192
191
190
189
^
34242
34439
34635
34830
35026
36218
36411
35(m
35793
36984
188
188
187
186
185
184
184
188
182
181
23
36173
36361
36549
36736
36922
37107
372<)1
37476
37668
37840
181
180
179
178
178
177
176
176
175
174
24
38021
38202
38382
38561
38739
38917
39094
39270
39445
39620
173
173
172
171
171
170
169
169
168
167
25
39794
39967
40140
40312
40483
40654
40824
40993
41162
41330
167
166
166
165
165
164
163
162
162
161
26
41497
41664
41830
41996
42160
42325
42488
42651
42813
42976
161
160
159
159
158
158
157
156
156
166
27
43136
43297
43457
43616
43775
43933
44091
44248
44404
446()0
165
154
154
153
152
152
151
151
151
150
28
44716
44871
45025
45179
45332
46484
46637
46788
46939
46090
149
149
149
148
147
147
147
146
145
145
29
46240
46389
46538
46687
46835
46982
47129
47276
47422
47667
145
144
143
143
143
142
142
141
141
140
SO
47712
47857
48001
48144
48287
48430
48572
48714
48865
48996
140
189
139
189
138
138
137
187
136
136
31
49136
49276
49415
49554
49693
49831
49969
50106
50243
50379
136
135
134
i;34
183
133
133
182
132
131
32
60615
50651
50786
50^)20
51055
51188
61322
51456
61587
51720
131
181
130
180
129
129
129
129
128
128
33
61851
51983
52114
52244
52375
62504
52634
52763
52892
53020
127
127
126
126
126
126
125
125
125
124
34
63148
53275
53403
128
53529
128
53656
53782
122
53S)08
54033
121
64158
54283
124
128
128
122
121
121
35
54407
54531
54054
54777
54900
55023
65145
55267
56388
55509
121
120
120
119
119
119
119
118
118
117
36
65630
55751
55871
55991
56110
56229
56348
56467
66586
56703
117
117
117
116
116
116
115
115
115
114
37
66820
66937
67054
57171
57287
57403
57519
57634
57749
67864
114
114
114
118
113
113
112
112
112
112
38
67978
58092
58206
68320
68433
58546
68659
68771
68883
58995
112
111
110
no
110
110
109
109
109
109
39
59106
59218
59329
59439
59560
59660
69770
69879
69988
60097
319
TABLE OF LOGARITHMS.
N
40
41
42
43
44
46
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
04
65
66
67
68
(59
O
1
2
3
4
6
6
7
8
9
108
(30206
106
61278
108
62325
101
63347
99
64315
108
60314
106
61384
108
62428
101
63448
98
64444
108
60423
105
61490
108
62531
101
63548
98
64542
107
60531
105
61595
108
62634
100
63649
98
64640
107
60638
105
61700
102
62737
100
63749
98
64738
107
60746
105
61805
102
62839
100
63849
97
64836
107
60853
105
61909
102
62941
99
63949
97
64933
107
60959
104
62014
102
63043
99
64048
97
65031
106
61066
104
62118
' 102
63144
99
64147
97
66128
106
61172
104
«2221
101
63246
99
64246
96
65225
96
65321
94
66276
92
67210
90
68124
88
69020
96
65418
94
66370
92
67302
90
68215
88
69108
96
65514
94
66464
92
67394
90
68305
88
69197
96
65610
94
66558
92
67486
90
68395
88
69285
95
65706
94
66652
91
67578
89
68485
87
69373
95
65801
94
66745
91
67669
89
68574
87
69461
95
66896
93
66839
91
67761
89
68664
87
69548
95
65992
93
66932
91
67852
89
68763
87
69636
95
66087
98
67025
91
67943
89
68842
87
69723
95
66181
98
67117
90
68034
89
68931
87
69810
87
69897
85
70757
88
71600
82
72428
80
73239
86
69984
85
70842
88
71684
82
72509
80
73320
86
70070
85
70927
88
71767
82
72691
80
73400
86
70157
85
71012
88
71850
81
72673
80
73480
86
70243
84
71096
83
71933
81
72754
80
73560
86
70329
84
71181
83
72016
81
72835
80
73640
86
70415
84
71265
82
72099
81
72916
79
73719
85
70501
84
71349
82
72181
81
72997
79
73799
85
70586
84
71433
82
72263
81
73078
79
73878
78
74663
77
75435
75
76193
74
76938
77670
86
70672
84
71617
82
72346
81
73159
79
73957
79
74036
77
74819
76
76687
75
76343
74
77086
79
74115
77
74896
76
75664
75
76418
74
77169
79
74194
77
74974
76
75740
75
76492
78
77232
78
74273
77
75051
76
75815
74
76667
73
77305
78
74351
77
76128
76
75891
74
76641
78
77379
78
74429
77
75206
75
76967
74
76716
78
77452
78
74607
77
75282
75
76042
74
76790
78
77525
78
74686
77
75358
75
76118
74
76864
78
77697
78
74741
76
75511
75
76268
74
77012
72
77743
72
77816
71
78633
70
79239
69
79934
^s
80618
72
77887
71
78604
70
79309
69
80003
68
80686
72
77960
71
78675
70
79379
69
80072
68
80754
72
78032
71
78746
70
79449
69
80140
67
80821
72
78104
71
78817
70
79518
69
80209
67
80889
72
78176
71
78888
70
79688
68
80277
67
80966
72
78247
70
T8958
69
79657
68
80346
67
81023
71
78319
70
79029
69
79727
68
80414
67
81090
71
78390
70
79099
69
79796
68
80482
67
81168
71
78462
70
79169
69
79865
68
80550
67
81224
(57
81291
66
81954
65
82607
64
83251
68
838a5
67
81368
66
82020
65
82672
04
83315
63
83948
67
81425
66
82086
65
82737
64
83378
63
84011
66
81491
65
82151
64
82802
64
83442
6:3
84073
66
81658
65
82217
64
82866
63
83506
63
84136
66
81624
65
82282
64
82930
63
83569
62
84198
66
81690
66
82347
64
82995
63
83632
62
84261
66
81757
65
82413
64
83069
63
83696
62
84323
66
81823
65
82478
64
83123
68
83759
62
84386
66
81889
65
82543
64
83187
63
83822
62
84448
320
TABLE OF LOGARITHMS.
N
70
71
72
73
74
76
76
77
78
79
80
81
82
83
84
86
86
87
88
89
90
91
92
93
94
96
96
97
98
99
O
1
2
3
4
5
6
7
8
9
62
W510
61
85126
60
85733
59
86332
59
86923
62
84572
61
85187
60
85794
59
86392
59
86982
62
84634
61
85248
60
85854
59
86451
58
87040
62
84696
61
85309
m
85914
59
86510
58
87099
62
84757
61
85370
60
85974
59
86570
58
87157
62
84819
61
85431
60
86034
59
86629
58
87216
61
84880
61
85491
60
86094
59
86688
58
87274
61
84942
61
85552
60
86153
59
86747
58
87^32
61
86003
60
85612
60
86213
59
86806
58
87390
61
86065
60
86673
•60
86273
59
86864
58
87448
58
87506
57
88081
56
88649
56
89209
55
89763
58
87564
57
88138
56
88705
56
89265
55
89818
58
87622
57
88195
56
88762
56
89321
55
89873
58
87679
57
88252
56
88818
55
89376
55
89927
58
87737
57
88309
56
88874
55
89432
55
89982
57
87795
57
8836(>
56
88930
55
89487
90037
54
90580
53
91116
58
91645
52
92169
51
92686
57
87852
57
88423
56
88986
55
89542
55
90091
57
87910
57
88480
56
89042
55
89597
54
90146
64
90687
58
91222
52
91751
52
92273
51
92788
51
932<)8
50
93802
49
94300
49
94792
48
95279
57
87967
57
88536
56
89098
55
89653
54
90200
57
88024
56
88693
56
89154
55
89708
54
90255
90309
54
90849
53
91381
52
91908
52
92428
54
90363
54
90902
53
91434
52
91960
52
92480
54
90417
53
9095()
91487
52
92012
52
92531
54
90472
5;^
91009
5;^
91540
52
92065
51
92583
54
90526
91062
53
91593
52
92117
51
J)2634
51
93146
60
93()51
50
94151
49
94645
49
95134
54
90634
53
91169
&S
91698
52
92221
51
92737
54
90741
91276
52
91803
52
92324
51
92840
54
90795
58
91328
52
91855
52
92376
51
92891
51
92942
50
93450
50
93952
49
94448
49
94939
51
92993
50
93500
60
94002
49
94498
49
94988
51
93044
50
93551
50
94052
49
94547
49
95036
51
93095
50
93601
50
94101
49
94596
49
95085
51
93197
50
93702
50
94201
49
94694
48
95182
51
93247
50
93752
50
94250
49
94743
48
95231
51
93349
50
93852
49
94349
49
94841
48
95328
51
93399
50
93902
49
94399
49
94890
48
95376
48
95424
48
95904
47
96379
47
96848
46
97313
46
97772
45
98227
45
98677
44
99123
44
99564
48
95472
48
95952
47
96426
47
96895
46
97359
48
95521
48
95999
47
96473
47
96942
46
97405
48
95569
48
96047
47
96520
47
96988
46
97451
48
95617
47
96095
47
96567
46
97035
46
97497
48
95665
47
96142
47
96614
46
97081
46
97543
45
98000
45
98453
45
98900
44
99344
44
99782
48
95713
47
96190
47
96661
46
97128
46
97589
48
95761
47
96237
47
96708
46
97174
46
97635
48
95809
47
96284
47
96755
46
97220
46
97681
48
95866
47
96332
47
96802
46
97267
46
97727
46
97818
45
98272
45
98722
44
99167
44
99607
46
97864
45
98318
45
98767
44
99211
44
99651
46
97909
45
98363
45
98811
44
99255
44
99695
46
97955
45
98408
45
98856
44
99300
44
99739
45
98046
45
98498
44
98945
44
99388
44
99826
45
98091
45
98543
44
98989
44
99432
44
99870
45
98137
45
98588
44
99034
44
99476
44
99913
45
98182
45
98632
44
99078
44
99520
43
99957
321
WHITB'S ALO. 21
822 ALGEBRA. [§612.
APPLICATIONS TO NUMERICAL PROCESSES.
1. Find the continued product of 875.43, 3.1416, and .00643.
log 875.43 = 2.04222
log 3.1416 = 0.49716
log .00543 = 3.73480
Sum = 1.17417 = log 14.933, Ans. (§ 690)
2. Find the quotient of 72.9 by 645.37.
log 72.9 = 1.86273
log 645.37 = 2.73670
Difference = 1.12603 = log .13367. (§ 691)
3. Find the 60th power of 1.06.
log 1.06 = 0.02119
60
Product = 1.06950 = log 11.468 .... (§ 692)
.'. 11.468 is the 60th power of 1.06.
4. Find the 12th root of 366.
log 366 =2.66229
Dividing by 12  2.66229
Quotient = 0.21362 = log 1.6350 •••. (§ 693)
5. Find the 6th root of .0366.
log .0366 = 2.66229.
Add to log  3 + 3, 5 + 3.56229.
Dividing by 5 (§ 593), 1.71246.
The antilog of 1.71246 = .61674, the required root.
612. When, as in Example 6, we have to divide a negative
characteristic which is not an exact multiple of the divisor, we
add to the negative characteristic as many negative units as
will make it such a multiple, and prefix the same number of
positive units to the mantissa. Thus, above, 2.66229 =
6 + 3.66229, and, dividing by 6, we obtain 1.71246, the antilog
of which is .61674, the required root.
§ 613.] LOGARITHMS. 823
When negative numbers occur in computations, we proceed
as if all were positive, deciding the sign of the result according
to the rule of signs. Thus, to obtain the product of 53.6 and
— 3.975 by logarithms, we proceed as if both were positive,
and give the negative sign to the result.
Find by logarithms the values of the following expressions :
6. 535 X. 342. ^^ .5673x25.05
7. 2.937x1.505. 763x365,47
8. 6.7354 X .8925. ^^' (^•^)'
9. 12345 X .0053782. ^^' (1.0546)'®.
10. .075 X .3678 ^ 73.251. ^® (^054 x .53798)*
56.32 8275.5 ^'^' V§J^
' 9.365* * 43296* 18. a/1.0562.
613. Exponential equations (§ 569) are most readily solved
by the aid of logarithms. When, for example, the a, I, r, of a
geometrical progression, are given to find n, the value of n
must be found from the equation lz=ar^~^] whence r*~' = Z 5 a.
Taking the logarithms of both members of this equation,
(n — 1) log r = log I — log a •
whence n = 1 + ^^S I log a .
logr
By taking from the table the logarithms of the given quan
tities a, I, r, and performing the indicated operations, we obtain
the value of n. If in the foregoing formula we suppose
a = 3, 1 = .00019683, r = .3, we obtain
n:=l \ ^Qg 00019683  log 3
'^l i^^:3 '
^ , 4.29409  0.47712
or =1H = ;
1.47712
whence » = 1 + §1^= 1 + 8 = 9.
1.47713
824 ALGEBRA. [§ 614.
19. Given a = 162, i= ^, r = J: find n.
20. Given a = ^, « = 4o, r = 4: find n.
21. Given a = 343, « = 400, / =  : find n.
22. Given r = 5, «=1562, Z = 1250: find ».
Solve the exponential equations
23. 11« = 1331. 26. 13' = 14. 29. 4'« = 6.
24. 5** = 15625. 27. 17' = 91. 30. 3'+* = 2187.
25. 12' = 41.57. 28. alf = c. 31. 3''^+* = 1200.
(iS^ee also § 689.)
BUSINESS FORMULAS.
614. Problems in percentage can in most cases be readily
solved by the methods of arithmetic ; but even in arithmetic
there may be advantage in the use of general formulas to in
dicate the processes involved.* Such formulas have the
advantage of being so related that they can be derived from
each other, and thus be readily reproduced when needed. In
algebra the solutions of all percentage problems may be indi
cated by formulas with the further advantage that they permit
the easy use of logarithms, thus greatly facilitating computa
tion.
Formulas for the more common processes in percentage, in
cluding simple interest, have already been given (§§ 285289);
and all that is further needed is the presentation of general
formulas for compound interest and annuities, with examples
illustrating the use of logarithms.
Compound Interest.
616. The formula for finding the amount of any given
principal at compound interest for any time and any rate per
cent may be obtained as follows :
Let p denote the principal, r the rate per cent, and n the time
* For methods of using percentage formulas in arithmetic, see Whitens
New Complete Arithmetic.
§ 616.] LOGARITHMS. 825
in years. Since the amount of one dollar in one year is 1 + r,
the amount of p dollars in one year is pil^r) ; and the amount
of p (1 h r) dollars in one year is
i)(H r) X (1 + r)=:i)(l + ^)',
which is accordingly the amount of p dollars in two years.
The amount of p(l {■rf dollars in one year is
p(l + r)«x(l + r)=i>(l + ^)',
which is the amount of p dollars in three years; and so on.
Hence the amount of p dollars in n years is p(l + r)" ; that is,
denoting the amount by a,
a=p(lfr)». (B^
616. Any three of the four numbers denoted by a, p, r, w,
being given, the fourth may be found from the above formula,
which therefore suffices for the solution of all cases in com
pound interest.
If a, p, r, are given to find n, we have to solve the exponential
equation
whence n log (1 + r) = log a — log jp,
and n = (log a — logp) ^ log (1 + r).
1. In how many years will $ 365 amount to $ 500 at 4 %
compound interest ?
Here a = 600, p = 365, and 1 + r = 1.64. Substituting the logarithm
of these numbers for those of a, p, 1 + r, above, we find
n =(2.69897  2.66229) r .01703 = 8.026.
2. At what rate per cent will $ 1500 at compound interest
amount to $ 1889.568 in 3 years ?
3. What principal will amount to $ 535.9572 in 3 years at
6 % compound interest ?
4. Find a formula to show how long it will take p dollars to
amount to mp dollars at r % compound interest.
326 ALGEBRA. [§ 617.
617. If the interest is payable at shorter intervals than
a year, say every half year, then, at r % per annnm, the
interest on one dollar for half a year is  ; and at compound
interest the amount of p dollars in n years, that is, in 2 n half
years, is jp[l+^] , the same as the amount of p dollars in 2 n
years at ^%. In the same way, if interest is payable m
times a year, the amount of p dollars at r % is expressed by
the formula
(• ^0
Ankuities.
618. To find the amount of an annuity accumulating any
number of years, allowing compound interest,*
Let a denote the number of dollars in the annuity, n the
number of years, 1 4 r the amount of a dollar for one year,
and A the required amount. At the end of the first year a is
due, and amounts at the end of the second year to a (1 + r) ;
hence the whole sum due at the end of the second year is
a h a(l I r) = a[l f (1 f r)]. At the end of the third year the
sum due is, in like manner,
a[l + (lhr) + (l4r)(l + r)] = a[l + (l + r) + (l + r)«].
By proceeding in this way, we find that the amount due at the
end of n years is
a[l + (l + r) + (l + r)^ + (l + r)»+...+ (l + r)^],
a G. P. whose first term is a, and ratio 1 + r ; hence we
obtain A = a ^] +^)'* "} =  [(1 + rY  11.
* Annuities at simple interest have practically no existence.
§ 620.] LOGARITHMS. 827
619. To find the present value of an annuity to continue a
certain number of years, allowing compound interest,
Let p denote the present value. The amount of p in n years,
that is, p(l f r)", should be equal to the accumulated amount of
the annuity in the same time ; hence
^=:('(iir) <"
If we suppose n to be indefinitely great, that is, the annu
ity to be perpetual, then — — becomes — =0, and the forego
ing formula becomes p = — (2)
620. Besides problems relating to annuities proper, these
formulas may be applied to the solution of a great variety of
problems concerning life insurance, value of estates, increase of
capital, population, etc.
Problbms.
1. How much should a man pay down to obtain a life
annuity of $ 1000, his expectation of life being estimated at 20
years, and interest reckoned at 5%?
Here a = 1000, n = 20, r = .05. Substituting these values for o, n, r,
in Formula (1), we obtain
_1000/j 1 \
^ .05 V (1.05)»A
By the aid of logarithms we find ^ = .3769... ;
hence p = ^522(i _ .3769) = 12462. Ans. $ 12462.
.05
2. What is the value of a farm that yields a net rental of
$ 900, interest reckoned at 6%?
328 ALGEBRA. [§ 620.
Here we hftve to find the present yalue of a perpetoal annuity of $ 900.
By Formula (2) above, we obtain
900 __ jg^jQ^ ^^ $15000.
^ .06 •
8. What is the present value of an annuity of $ 600 for 25
years, interest reckoned at 5%?
4. What sum should be paid for a 15 years' lease of a prop
erty yielding a net annual profit of $ 500, interest at 6% ?
6. By giving up smoking, a man saves $ 50 a year. How
much does he thus save in 30 years, interest at 5% ?
6. A man with a capital of $ 10,000 spends every year $ 150
more than his income. In how many years will his capital be
consumed, interest reckoned at 8%?
7. A population of 1,000,000 has a steady annual increase of
3%. In how many years will it double itself ?
8. A man with a capital of $ 20,000 adds to it yearly $ 200
besides the interest. What will be the amount of his capital
in 20 years, interest reckoned at 10%?
9. Having borrowed $12,000 at 5%, how much should
I pay every year so as to discharge the debt in 10 years ?
10. In order to accumulate $ 3000, to cover the expenses of
a son's college course, a father invested a certain sum at 5%,
on each recurring anniversary of his son's birth until the 18th
inclusive. How much did he invest yearly ?
11. What annual payment will amount to $5000 in 21
years at 6% compound interest ?
12. What is the value of an estate yielding $1500 net
annual income, interest reckoned at 5^%?
§622.] UNDETERMINED COEFFICIENTS. 329
CHAPTER XXI.*
UNDETERMINED COEFFICIENTS AND APPLICATIONS.
621. It is sometimes desirable to obtain for a given alge
braic expression an equivalent of a certain form ; as, for ex
ample, a series in ascending powers of x. To this end, we
first assume that the given expression is equal. to a series of
the required form, but having undetermined coefficients. We
then proceed to determine the values of these coefficients by
means of the principles exemplified below.
622. (1) If an equation of the form
is such that for every value of x the equation is an identity
(§ 624), then the coefficients of the like powers of x in the two
members are equal.
For, since the equation is satisfied by any value of «, let
a; = 0. Then every term containing x equals 0, and the equa
tion reduces to
Subtracting these equals from the members of Equation (1),
we have
Bx+C7?^D7?'{ ... = B'x 4 (7'aj2  D'oi? + ... (2)
Dividing each member of (2) by x, we have
JB + Ca: + Z>a^ f  = ^' + O'a; + D'x" + .... (3)
* This and the following chapters are designed for more advanced
classes, and especially for stiMents who are preparing for higher institu
tions who^e entrance requirements may include an elementary knowledge
of one or more of the subjects treated.
330 ALGEBRA. [§ 623.
Since Equation (3) must be satisfied by any value of a?, let
a; = ; and then the equation reduces to
In the same way we can prove that C = C, D=D\ and so on.
623. (2) If an equation of the form
^h5a?hOc' + l>«'+ — =0 (4)
is true for every value of a?, the coefficients must each eqtuil zero.
For, since x may have any value, let a? = 0, and then the
equation reduces to
^ = 0.
Then omitting A in (4), and dividing by a?, we have
J5 + (7a; + 2>ar*+... =0. (5)
Since x may have any value in (5), let a? = 0, and then we
obtain B = ; and in the same way we can prove that C7= 0,
D = 0, and so on.
624. It can be shown that the foregoing principles (§§ 622,
623) apply only to equations that contain convergent series
(§ 631, note), but the proof belongs to higher algebra.
Let us now consider several of the more important applica
tions of these principles.
RESOLUTION OF FRACTIONS.
625. To resolve a fraction into partial fractions is to find the
fractions whose algebraic sum is the given fraction. This is
the converse of the process of adding fractions given in § 264
For example, ^ ^ = ^^ r^; and conversely,
^ x — b a;f3 or — 2aj — 15
the fraction „ ^'^^^ ^^ being given, it is required to find
a^ — 2 a; —• 15
the partial fractions of which this given fraction is the alge
braic sum.
§626.] UNDETERMINED COEFFICIENTS. 331
Having factored the denominator, we assume that
x'^2x15 x + 3 x5'
A and B being the numbers which we wish to determine.
Clearing (1) of fractions, we obtain
x + 19 = Ax5A + Bx + 3B,
or a; 4 19 =(^ f B)x(5A  SB).
Equating the coefficients of the like powers of x (§ 622), we
obtain
A + B=l, (2)
5J.35 = 19; (3)
two equations each containing A and B, from which we obtain
^ = 2, 5 = 3;
whence — = •
ar^2aj15 x5 flj + 3
626. Suppose, however, that it be required to resolve into
partial fractions the expression ^ ~ ^^«
or — 2 X — 15
If, having factored the denominator, we should assume that
2«213a;9 A B
x'2x15 a; + 3 aj5'
we would be led, on clearing of fractions, to the absurd result
2 = 0. This difficulty may be avoided by first reducing the
fraction — — ^^^ — ^"^ to a mixed number, and then resolving
or — 2 X — 15
the fractional part, as shown below.
Hence, whenever the numerator of the fraction to be resolved
is not of lower degree than the denominator, reduce the given
fraction to a mixed number, and then resolve the fractional part.
332 ALGEBRA [§ 627.
Thus, taking the fraction given above,
. 9xH21 A . B
Assume — ^ — • = \
(a; + 3)(a;5)~aj3 x5'
then, clearing of fractions,
■ 9 X \21 = Ax 5 A\Bx\ SB = (A^B)x (5 A — 3 B) ;
whence, equating, A^ B = — 9] 5^ — 35 = — 21;
whence A = — 6, 5 = — 3;
whence H^zl^^^^2 ^ ^
a^2x15 x\3 x5
Resolve into partial fractions
2a;13 , 2a^ + g43
aj2 _ 13 aj f 40' ' a^1
2a;415 « 3a^43a^f2
ar^ — 15 ic + 56 a? — x
7a?23 ^ 2 a^ f 21 a; + 13
ar«_6a; + 5" ' 4a^5a^hl '
4a? 29 jQ 3a^ + 3a;Hi8
ar^ + 3a;10* ' a^9x
13 a;+g ji 8a; 112
6a^^5x\l' ' a^\6x\S
2a^ + 2x6 ,„ 2«2 + a:l
aj34.5aj24.6aj 2a;2^^_3
EXPANSION OF FRACTIONS INTO SERIES.
627. (1) Let ^^^^ ^ be the given fraction.
1 — a? — ar
Assume ."^"^^^ , = A{ Bx\ Ca^ { Di^ { '" (1)
1 — a; — ar
1.
2.
3.
4.
5.
6.
§628.] UNDETERMINED COEFFICIENTS. 333
Multiplying each member of (l)hj 1 — x — oi^, we obtain
l + 2x = A^B x\C a^iD aj»+etc.
A
x\C a^iD
B
0
A
B
Whence A = l', BA = 2, .. ^ = 3; (7(^ + J5)=0,
.. 0=4; D(5+O)=0, .. i> = 7; etc.
Substituting these values in the assumed series (1), we
obtain ^i±l^ = l + 3a; + 4ic2 + 7ar^ + ll«*418a^+..., in
1 — a? — ar
which each coefficient after the second is equal to the sum
of the preceding two coefficients, and hence the series can be
continued indefinitely.
628. (2) Let — ^^ — ^^ be the given fraction.
^^ aj23a;2 ^
We see by inspection that the first term of the development
must contain x~^\ and we accordingly assume that
^^^ =^2 + J5a;i + + i>a+ — , (1)
aj*3a?h2
and proceed as above.
Similarly, in other cases, we should first determine by in
spection what power of x must occur in the first term of the
development, and then assume the series accordingly.
1 — 2r 1— 2rr
The fraction —  — ==^ — may be written in the form — , and
then developed as in (1) in § 627.
Such developments as those here referred to may be obtained by three
distinct methods: to wit, (1) by simple division^ which may be em
ployed as a means of proving the correctness of the results obtained
by other methods; (2) by the method of undetermined coefficients^ as
exemplified above; and (3) by means of the binomial formula ^ as
shown §§ 630, 631.
384 ALGEBRA* [§ 629.
Expand into an infinite series
1. UM^. 6. '
2. A 1 6.
l+3a;
1
1
2aj
+*•
1
X
1
2a;
3a?
1 +a;
3. ^ 1. — n__.. 7.
4 . "^" 8.
1aj + a?
BINOMIAL FORMULA.
3a;
l\x
l2xa?
14 a;
l+2a;43a;*
ahbx
l+2a; + 3a?
629. It has been shown by successive multiplications of the
binomial a + & (§ 321), that, if n denotes any positive integer
from 2 to 5 inclusive,
(a + by = a* + na^'b + '^^'^ "J^^ a^^'b'
1 • ^
, w(n — l)(n — 2) ^_3,3 , ,^.
+ ^ 1.2.3 ^« ^M— •. (1)
It remains to prove that this formula holds true for any
positive integral value of n (§ 324).
Assume that Formula (1) is true for any positive integral
value of w. Multiplying both members of (1) by a 4 6, we have
(l)x(4&) a»64 na»^624 !L^:^Da"V4— •
1 • i6
Collecting terms, we obtain
(a 4 6)""^' = a"'' 4 (w 4 1) a^h + i^±I)^aW
1 • 2
§630.] UNDETERMINED COEFFICIENTS. 886
Comparing (1) and (2), we see that the same law holds in
both for the formation of the coefficients, and also for the
exponents ; and hence, if the formula holds true for the expan
sion of (a f by, it holds true for the expansion of (a 4 by^\
But it has been shown by actual multiplication that Formula
(1) holds true when n = 5 (§ 321), and hence it holds true
when n = 6 ; and, if it holds true when n = 6, it holds true
when n = 7, and so on indefinitely to any power denoted by
a positive integral exponent
Again, if 6 is a negative number, then, since the odd powers
of a negative number are negative, and the even powers positive
(§ 318), we evidently have, from (1),
(a + by = a»  na^^b + ^^^"^^ a^'y
w(n — l)(n — 2) ^ .,3 , ,^.
1.2.3 ^a* '&* + —, (3)
the alternate signs being f and — .
Equations (1) and (3) can be written as one formula as follows :
(a ± by = a ± wa«*6 + ^^^ "^''"^ «""^^'
1 • 2
^ 1.2.3 — ^* ■*■ ***• ^^
Negative or Fractional Exponents.
680. It can be shown that the above formula (A) holds also
for both negative and fractional exponents; but the results
obtained are reliable only when the resulting infinite series is
convergent.
When the exponent n is a positive integer, the series ter
minates with n + 1 terms ; for the coefficient of the next and
each succeeding term contains the zero factor n — n, which
causes the terms to vanish. But when n is negative or frac
tional, no factor can become zero, and hence the series will
386 ALGEBRA. [§ 631.
never terminate, but is an infinite series, either convergent or
divergent.
631. The eonvergency and divergency of series is too diffi
cult a subject for consideration in an elementary algebra, and
the same is true of a perfectly rigorous demonstration of the
binomial formula for negative and fractional exponents.*
* None of the proofs given in the ordinary algebras are free from
objections ; and only a few of the more recent make any reference to
failure of the formula in the case of divergent infinite series.
We here add a few notes respecting the eonvergency and divergency
of series to indicate the difficulty of the subject.
A series consisting of an infinite number of terms, which succeed each
other according to some fixed law, is said to be convergent when the sum
of its first n terms approaches nearer and nearer to a finite limiting
value, according as n is taken greater and greater; and this limiting
value is called the sum of the series, and from this value the series can
be made to differ by an amount less than any assigned quantity on taking
a sufficient number of terms.
If the sum of the first n terms approximates to no finite limit, the
series is said to be divergent.
Let us illustrate by a few examples.
Take J— = lhx + x^\x^ + ....
1 — X
If X < 1, this series approaches a fixed limit. If, for example, x = J,
then we have
' =3^1 + 1+1+1+....
1  J 2 3 32 38
The greater the number of terms taken, the nearer the sum approaches
the fixed limit , and similarly for any value of x less than 1. Hence
the series is convergent if x < 1.
On the other hand, if x > 1, the series is divergent. For example, let
a; = 2, and then we have the manifestly absurd result
J— = 1=1 + 2 + 4 + 8 + ....
12
In this case the sum does not approach a fixed limit, and the series is
therefore divergent:
Again, let —1— = l  x + x^  sfi + ....
1 +»
§631.] UNDETERMINED COEFFICIENTS 337
We shall therefore assume the formula to be true for such
exponents in cases in which the values of a and b are such as
to result in convergent series; and, in the examples given
below, care will be taken to indicate when the resulting series
is convergent.
1. Expand (a + h)'\
Here n=l ; ^^ =^:i^ = l ; ?^ = ' 1; ?^ = i±^ = l.
It is thus seen that the coefficients are alternately + 1 and — 1, since
the product of an even number of negative factors is +« a^d of an odd
number of factors — (§ 106).
Hence
(a + 6)i = ai  a^b + a^^b^ ... + a'6'i = (1 + ^+),
o \ a Qi Cb I
a series which is convergent when a > &.
It is often found convenient, as in this example, to place a factor out
side a parenthesis, and to change negative to positive exponents.
2. Expand (a + &)*
Using formula A^ b being positive, we have
(af 5)i=ai+ 1 a^b ^ ah^^ ^ ah^ ^^^ a'hi^...
V 2a 2. 4a2 2.4. 6a8 2.4.6.8a* /
This series is also convergent when a>b.
It is usually well, as in this series, to keep the factors of the coefficients
separate, so as to show what is termed the law of the series.
Here, if x < 1, the sum of the series approaches a fixed limif, and
hence is convergent.
If a; = 1, then the series becomes 1 — 1 + 1 — 1 + •••. For an even
number of terms the sum is ; but for an odd number of terms the sum
is + 1. In this case the sum oscillates from to + 1 without approach
ing a fixed limit, and hence the series is divergent.
If X > 1, the sum of an even number of terms of the series is always
negative, and the sum of an odd number of terms is always positive.
Hence, in this case, the sum does not approach a fixed limit, and the
series is therefore divergent.
WBITS'S ALO. — 22
888 ALGEBRA. [§ 632
8. Expand or (a f &) .
Here«J,y^ 4' 3 " 6'^" g,aiia800ii
Hence (a + 6)"* = a"*  5 «"' & + ^^ «"**^  ^"4^ «"^^» + 
V^V 2o 2.4a« 2.4.6a« J*
This series is convergent when a > 6.
Expand to four terms
4. (1 + iV)'* 7. (2 + i)* 10. (a + 6)«
6. (1i)"*. 8. (1+a)* 11. (a6)~*.
6. (1+i)* 9. (la)l 12. (1 + i)*.
The series in 8 and 9 is convergent when a < 1, and in 10 and 11
when a > &.
Extraction op Roots by the Binomial Formula.
632. The binomial formula may be employed for the extrac
tion of any roots of a given number N.
For ^^= Va**±6 = a[l ± — )«, a* being, the nth powei
which is near to N, either greater or less than N\ but — must
a"
be a proper fraction, otherwise the expansion will be a
divergent series, and the result will not give the correct root.
When (1 )» is expanded, the number of terms required
a"
to give a certain degree of approximation will depend upon the
relative value of a** and 6.
As an example, let us find the square root of 72.
V72 = V64+8 = 8 (1 + 1)*, since ; = ^ = .
§633.] UNDETERMINED COEFFICIENTS. 889
Expanding (1 + \y by Formula (A), we have (since a = 1,
b = l, and n = ^),
\ Sy V 2 8 2.48' 2.4.6.
8»
1.3.6
+
•)
2.4.6.8.8*
Performing the operations indicated^ we obtain
V72 = 8.48528,
which is correct to the fifth decimal figure.
But few terms are necessary in this example, since a" is 8 times 6, and
the convergence of the series is rapid.
1. Extract the cube root of 17.
Here the third power nearest to 17 is 8, whence
v^ = y/sT^^ 2 (1 + f)i.
But since ^ is an improper fraction, the expansion of (1 + )^ wi]\
give a divergent series^ and the result will not be a correct value of ^17.
But since y/vi = y/27  10 = 3 ^1  J?, we can write
^=3fi15\* = 3(i^l?^fl5Vlilfl5V
V 27/ V 3 27 3.6\27y 3.6.9^27^
2.6.8 /loy \
3.6.9.12\27/ ** /'
This is a convergent series, and, by taking a sufficient number of terms,
we can find the value of vTf to any required approximation. The con
vergence of the series is, however, not rapid.
633. The following examples will converge rapidly, and but
a few terms will be required to obtain results correct to five or
six places of decimals.
2. ■\/2S. 3. a/34. 4. \/i29. 6. </244.
This process of extracting the roots of numbers, though interesting as
an application of the binomial formula, is practically of little value when
logarithmic tables are available.
840 ALGEBRA. [§631
CHAPTER XXII.
DETERMINANTS.
634. An expression whose value depends on two or more
quantities is said to be a function of those quantities. Thus,
a6 h 6c is a function of a, 6, and c (§ 640).
In solving a system of equations of the first degree, also
defined as linear, a class of functions appear, called deter
minants, first observed by Leibnitz in 1693.
635. Let us take the equations
faia; + 6,2^ = 0, (1)
(a2a; + % = 0. (2)
Multiplying (1) by h^ and (2) by — 6^ adding, and then
dividing the resulting equation by x, we have
ai62 — o,j^i =5 0.
The expression afi^ — ajb^ is a determinant It is usually
written with vertical lines at the left and right, called the
«gware/orw, thus: ^ ,\ Hence ^ ^ =0^2 — 0^1*
636. The first member of this identity is the undeveloped
form of the determinant, and the second member its developed
form.
The numbers ai, 02, 61, and b^ are called the elements of the
determinant, and the expressions 0162 and 0261 in the developed
form are called its terms.
The horizontal lines of letters in the undeveloped form of a
determinant are called rows, and the vertical lines of letters
are called columns.
S 639.]
DETERMINANTS.
341
The determinant
has two rows and two columns, and,
in the developed form (afiz — aj)i) each term is the product of
two factors ; hence the determinant is said to be of the second
order.
It will be seen later that in every determinant there are as many rows
as colunms, and as many of each as there are elements in each term.
637. It will be observed that the developed form of the
determinant
^2 62
is the product of the elements in the
diagonal from the upper lefthand corner to the lower right
hand comer (afi^ minus the product of the elements in the
diagonal from the lower lefthand corner to the upper right
hand comer (a^ bi).
The first of these diagonals (ai b^ is called the principcd
diagonal, and the second (0261) the secondary diagonal.
638. Since
Oi bi
02 ^2
= afii — djbi = afii — bia^ it follows that
bi 62
Hence the value of the determinant is not altered by
changing the rows into columns, and the columns into rows,
639. Again, since afi^ — ajbi = — (biO^ — ftjOi) = — (ajbi — arjb^,
it follows that
ttl 61
61 Oi
02 62
a, 62
&2 <h
Oi bi
Hence, if the order of the columns or rows be changed,
the sign of the determinant wiU be changed, but its absolute valve
will not be altered.
It will be seen later (§ 654) that the above laws hold true for a deter
minant of any order.
842
ALGEBRA.
r§640.
640. Let us now solve the simultaneous linear equations
(ai^ + bjy = Ci, (1)
( ojaj + 6^ = Cj. (2)
Multiplying (1) by &s and (2) by —hu and adding the resulting equa^
tions, we obtain
Multiplying (1) by — a,, and (2) by Oj, and adding the resulting equa
tions, we obtain
It will be observed that the numerators and the denominators in the
values of x and y are determinants, the denominators being alike, and
hence we can write
x =
Cl
6.
c.
h
«1
6.
0,
6.
(3)
V =
<h
<H
a,
c«
«i
&i
a,
ft.
(4)
Comparing the numerators and denominators, it will be seen
that the numerator of (3) can be obtained from the denomina
tor by changing the column of a's in the denominator to &s ;
and, in like manner, that the numerator of (4) can be obtained
from the denominator by changing the column of I/a in the
denominator to c's.
EXBRCISBS.
1. Solve, by means of (3) and (4), the equations
3aj + 4y=17.
Here ai = 5, as = 3 ; &i = 2, &2 = ^ ; ci = 19, and Ci = 17. Substitut
ing these values in (3) and (4), we obtain
{
« =
19 2
17 4
7634 o
= 206=^' y =
6 19
3 17
8667
6 2
3 4
6 2
3 4
~ 206
= 2.
§642.]
DETERMINANTS.
848
Solve in like maimer the following linear equations :
2.
3 (4ajy = 18,
\2x\3y = 16.
\5x2y = 3.
ax + by = c,
dx + cy =/.
5
•{
Expand the following determinants :
6.
5
4
3
2
7.
15
3
5
8.
4
3
5
DETERMINANTS OF THE THIRD ORDER.
641. Let us take the equations
aix 4 &iy + Ci« = 0, (1)
a^\b2y\c^ — 0, (2)
a^\b^{c^ = 0, (3)
MultiplyiDg (1) by 62C8  68C2, (2) by  biCz + bzCi, and (3) by 61C2
— b^u suid adding the resulting equations, thus eliminating y and z, and
then dividing the resulting equation by x, we obtain
ai(6aC8  &8Ca)  aaC&iCa  fesCi) + azipic^  b^i) = 0,
or
ai
62 Ca
— 02
61 Ci
+ fl8
61 Ci
68 Cs
bs cs
62 ca
= 0.
(4)
(6)
If the operations indicated in (4) be performed, we have
62 C2
02
61 Ci
+ 08
61 Ci
68 Cs
6« C8
62 C2
(6)
642. The first member of (6) is a determinant ; and, since
each term contains three demerUSj it is called a determinant of
the third order.
Such a determinant is usually written in the square form, and hence
(6) may be written
(7)
a\ 61 c\
62 C2
61 Ci
61 Ci
02 O2 C2
= 01
6s cs
Oa
6t ct
+ 08
b% 0%
Ot 08 Cg
344
ALGEBRA.
t§«4d.
«1
bi
Cl
a2
62
ca
«3
bz
cs
ai
bi
Cl
a2
b2
d
643. A determinant of the third order may be readily
developed as follows:
Repeat the first and second rows below, as at the right
For the positive terms, begin with m, «2» «8» respectively,
and multiply the elements diagonally downward, obtaining
€iib2Ci, ozbzCi, asbic^.
For the negative terms, begin with as, ai, 02* respectively,
and multiply the elements diagonally upward^ obtaining
ttsbifiij aibzc^i a2&iC8.
The development is aibfps + OzbsCi + asbii^ — azb2Ci — aibsC2 — aaftiCg.
In practice it will not be necessary to write down the repeated rows,
for the work can easily be done mentally.
644. If we change the rows to columns, and the columns to
rows, we shall obtain the same result. Thus,
ai 02 az
= aib2Cz + biC2as 4 CiOa&s — fliC2&8 — bia2Cs — CiboQz' (8)
61 62 bs
Cl C2 Cz
It thus appears that the value of a determinant of the third
order is not altered by cTianging the rows to columns, and the
columns to rows.
645. If the order of any two rows or columns be inter
changed, the result obtained will be the negative of (8).
Thus,
02 Oi as
&2 bi bz = 02&1C8 + 62Cifl8 + C2O168 — Ca^ids ~ b^fliCs — a^^bgci.
Ca Cl C8
It is thus seen, that, if any two columns or rows of a deter
minant of the third order be interchanged, ths sign of the
determinant will be changed, but its value will not be altered,
646. It appears from the first member of (6), § 641, that a
determinant of the third order consists of six terms, three of
which are positive, and three negative.
It is seen from (7) that a determinant of the third order
can be resolved into the sum of three determinants of the second
order.
647.]
DETERMINANTS.
345
EXBRCISBS.
Develop the following determinants :
1.
2.
3.
3
2
5
1
3
2
= 24.
4
1
3
5
1
4
3
2
5
=15. .
4
1
2
8
1
6
3
6
7
= 360.
4
9
2
7.
SI
iiow that
4.
5.
6.
8 4 5
2 3 1
5 2 6
1 2
2 6
4
a
b
c
a
1
2
1
3
2
3
4
8
3
2
2
=86.
2
4
1
=365a2c.
c
2
1
2 5
= what ?
4 3 2
5 16
647. Let us now solve the linear equations
(1)
(2)
(3)
Multiplying (1) by &2C8&8C2, (2) by ftiCs+ftsCi, and (3) by 61C262C1,
and adding the resulting equations, the terms that contain y and z vanish,
and we obtain
x =
dih^Cfi — d\bzC2 — d^biC^ + d2Jb%c\ .+ d^bic^ — ^3^2^!
.1. I i_   ..  , I ^
ai&2C8 — a\hzC2 — a^biCz + 02^8^1 + azhic^ — azb^\
(4)
Similarly, multiplying (1) by — 0^10% + azC2i (2) by aiCs  azCu and
(3) by — a\C2 4 «2Ci, and adding the resulting equations, the terms that
contain x and z vanish, and we have, after ieirranging the terms,
y =
01^2^8 — dxdzCj — a^dxCz + ciidzC\ + (izd\c^ — g8d2Ci ,
ai&flCs — ai&8C2 — 02&1C8 + oa&aCi + fls&iCa — as&aci
(6)
846
ALGEBRA.
[§647.
LasUy, multiplying (1) by a^&s — Os&a, (2) by — aibz + as&i, and (3)
by ai&s — fH^u and adding the resolting equations, the terms that con
tain X and y vanish, and we have, after rearranging the terms,
Z r= '
(6)
The numerator and the denominator in the above values of x, y^ and z
respectively are determinants of the third order, and they may be written
in square form, thus :
X =
di
bi
Cl
di
&2
C2
dz
bz
<58
«i
bi
Cl
0?
&2
C2
az
68
C8
(40 » =
ax
dx
Cl
Oi
d2
C2
az
dz
cz
«i.
bi
Cl
02
62
Ci
az
bz
Cz
(5')
z =
ai
&i
<il
02
&2
d2
az
68
dz
Ol
61
Cl
02
&2
Ct
as
&8
Cz
(60
It is seen that the denominators in these values of x, y^
and z are the same; and that the numerator of the value of
X can be obtained from its denominator by changing the column
of a's in the denominator to d's; that the numerator of the
value of y can be obtained from its denominator by changing
the column of &'s in the denominator to d's; and that the
numerator of the value of z can be obtained from its denomi
nator by changing the column of c's in the denominator to d's.
rSa; f4y22; = 17,
8. Solve the linear equations <5a;f ^ + 32 = 16,
Uajf 3y72 = 31.
Here ai = 3, 02 = 5, as = 4 ; 61 = 4, 6a = 1, 69= 3 ; Ci = 2, C2 = 3,
cg = 7 ; (?i = 17, ^2 = 16, dz = 31. Substituting these values in (4'), (6'),
and (6'), we obtain
X =
17
4
2
16
1
3
31
3
7
3
4
2
5
1
3
4
3
7
_ 119 + 96 + 372 ~ 62  448  163
21 + 30 + 48814027
76
76
= 1.
§648.]
DETERMINANTS.
847
y =
3
17
2
5
16
3
4
31
7
162
3
4
2
~ 76
5
1
3
4
3
7
= 2. z =
3
4
17
5
1
16
4
3
31
228
3
4
2
~ 76
6
1
3
4
3
7
= 3.
The values of x^ y^ and z may be written as determinants directly,
without substituting the values of the elements. The denominator in
each determinant should be written first, and then the numerator,
according to the direction given above.
Solve the following linear equations :
3aj — 2y— 2; = 4,
9. ^5aj3y+ 2; = 10,
2x + ^ySz = ll.
2a; 4^ + 32 = 10,
10. l 3x+ y — 2z = 6y
x + 3y— 2 = 20.
6a; — 3y+ 2
11. ^9x\2ySz
x — 4,y — 5z
x+ y+ z
12. ^ 5x^4:y^Sz
3a; 44^ — 32
= 16,
= 14,
= 10.
= 6,
= 22,
= 2.
DETERMINANTS OF ANY ORDER.
648. A determinant with n rows and n columns is called a
determinant of the nth. order. It contains n' elements.
A determinant of the nth order may be developed by taking
the sum, with the proper sign (to be established later), of all
the possible products of its w* elements that can be formed
by taking as factors one element, and only one, from each
row, and one, and only one, from each column. Each term
of the developed determinant will contain n elements
Take, for example, the determinant :
This is a determinant of the fourth
order ; and the sum of all possible prod
ucts of its 16 elements in sets of 4, one
being taken from each row and one from each column, will be its de
velopment or value.
ai
61
Cl
di
as
&2
Ca
d2
as
bs
Cs
ds
a4
64
Ca
^4
348
ALGEBRA.
t§649.
649. The product of the elements in the principal diagonal
from ai to ^4, which is afi^c^i, is taken as the leading term.
Its subscripts are in natural order from left to right (1, 2, 3, 4),
and it is regarded as positive.
The signs of the other terms are determined by the number
of interchanges required to bring the subscripts in each in their
natural order. If the number of interchanges is even, the
term is positive; if the number of interchanges is oddy the
term is negative.
Thus, in a2&iC4d8, to bring 1 to the first place, it must be interchanged
with 2, and to bring 3 to the third place, it must be interchanged with
4, making two interchanges, and hence the term is positive. But in
ai&4C8f?2, the 2 must be interchanged with 3 and with 4, and 3 must be
interchanged with 4, making three interchanges, and hence the term is
negative.
650. A determinant may be indicated by its leading term
in vertical lines. Thus,  % 62 1 denotes a determinant of
the second order ;  ai 62 C3 1, a determinant of the third
order ;  Oi ^2 Ps ^4 1> 3, determinant of the fourth order ; and
I tti &2 C3 c74 ••• r„ I, a determinant of the nth order. A deter
minant may also be expressed by the notation 2 ± before
its leading term, as 2 ± a^b^.
MINORS AND COFACTORS.
651. If we strike out of a determinant any number of rows
and the same number of columns, the remaining elements
form a determinant called a minor. This minor, and the de
terminant formed by the elements common to the rows and
columns stricken out, are called cofactors.
Take, for example, the determinant (1), and strike out the
first row and the first column as in (2). Then (3) is its first
minor, and Oi is its cofactor.
(1)
tti hi Ci di
CI2 bi Ci di
(2)
<tt &s Cs ds
04 &4 C4 (I4
fli &i Ci di
c^ h% C2 d^
as &8 cs dg
94 &4 C4 d4
(8)
&2 Cs ^2
63 Cs dz
(4)
64 C4 d^
653.]
DETERMINANTS.
349
If we strike out the first and second rows and the first and
second columns, as in (4), then
p8 ^
C4 ^4
its cof actor, or corresponding minor.
is the minor, and
652. From the determinant (1) above, we can obtain the
cofactors.
(1)
(2)
(3)
(4)
b2 C2 di
61 Ci di
61 ci di
61 Ci C?!
hs cs ds
02
bz cz ds
as
bi C2 dz
«4
&2 ^2 di
64 C4 d4
64 C4 d4
64 C4 di
bz Cz dz
ai
It will be seen that (1) contains all the possible terms that
contain ai ; (2), all the possible terms that contain Oj ; (3), all
the possible terms that contain Og; and (4), all the possible
terms that contain a^. But in each one of these minor deter
minants there are six terms, as shown in § 646, and hence
there are 24 terms (6 x 4) in a determinant of the fourth
order.
In like manner it can be shown that there are 120 terms, (24 x 5) or
(1x2x3x4x6), in a determinant of the fifth order ; and it may be
shown that the number of terms in a determmaut of the nth order is the
continued product of 1 x 2 x 3 x 4 ••• x n.
653. It further appears from the aboye that the expressions
Ol
taken together, comprise all the terms in the original deter
minant (1) in § 651.
It is evident that the first is positive, since, when each term of the minor
is multiplied by ai, the order of the subscripts is not changed. The
second is negative, since, when each term of the minor is multiplied
by 02, one interchange is necessary to bring a^ to the second place,
llie third is positive, since, when each term is multiplied by az, two
interchanges are necessary to bring az to the thii*d place. The fourth
is negative, since, when each term is multiplied by a^, three interchanges
are necessary to bring a^ to the fourth place.
bi C2 di
61 Ci di
61 ci d\
bi c\ d\
bz Cz dz
, a2
bz Cz dz
» «8
b^ C2 d^
, and ai
b^ C2 d^
64 C4 d4
&4 C4 di
64 C4 di
bz Cz dz
850
ALGEBRA.
[§654.
&2 Ci di
61 Cl <?1
61 Cl <Ji
61 Cl di
=ai
bz Cs dz
aa
&8 cs ds
+ a3
&2 Ca da
04
6a Ca da
64 C4 di
64 C4 d4
6$ Cs dg
63 Cs ds
Hence the following identity :
ai 61 Cl di
aa &a <Hi da
as bz Cs dg
a4 64 C4 d4
It is evident from the foregoing conaiderations that a determinant of
the nth order can be made to depend on n determinants of the order of
n1.
PROPERTIES.
654. Let us here recapitulate the properties of determinants
already established; and add a few others which are of special
value in manipulating determinants.
1. The value of a determinajit is not cUtered by changing the
roivs to columns, and the columns to rows.
This has been shown to be true of determinants of the second and third
orders (§§ 638, 643). It is also true of determinants of any order, since
the succession of terms is the same in the changed form as in the original
determinant.
2. If any two rows or columns be interchanged, the sign of the
determinant will be changed, but its value will not be altered.
This has also been shown to be true of determinants of the second and
third orders (§§639, 645). It is generally true; for, if the rows inter
changed be adjacent, the changed form will require one more interchange
than the original form to bring the subscripts into regular order ; and, if
the rows interchanged are not adjacent, an odd number of interchanges
will be necessary.
3. If two rows or columns of a determinant are identical, the
determinant vanishes.
For let D be the value of the determinant. Then, if the identical rows
be interchanged, the sign of the determinant will be changed by Property 2,
but its value will not be altered. Hence i>=—i>. .•. 2 2>=0. .*. Z>=0.
4. If each element in any row or column be multiplied by the
same number, the determinant will be multiplied by that number.
This is evident from the fact that every term of a determinant contains
one, and only one, element from the same row or column.
mai &i
Hence
ma% b%
= maibi — fna%bi = m(ai&a — a^i)*
§654.]
DETERMINANTS.
S51
5. If the elements of any row are like muUiplea of any other
row, the determinant vanishes.
For
mai
ai
bi
Ol
ai
bi
mai
a^
bi
= w
as
as
62
tnaz
as
68
as
a8
bs
= 0.
The determinant is put in the second form by Property 4, and this in
turn vanishes by Property 3.
6. A determinant of the nth order can he expressed in n deter
minants of the (n — l)th orderc
This has been proved in § 653.
7. If all the elements hut one of any row are zeros, the order of
the determinant is reduced hy one.
For if A\t Az, As, A4, etc., be the minors corresponding to au <hi as* 04,
etc., and if D denotes the determinant, then
D = aiAi — 02^2 + asAs — a^A^ + etc.
Now, let Di represent the determinant when all the elements au (h^
Qif 04, etc., except one (say ai), are equal to zero, then Di = aiAi. But
the order of 2>i, which is the same as that of Ai, is one less than D :
hence the property is proved.
8. If every element of any column or row is a sum of two or
Tftore numbers, the determinant is equal to the sum of two 07' more
determinants of the same order.
ai + fc + Z
6,
Cl
For
02 + wi + n 62 C2
08 + » + p 68 Cz
= (ai+* +
62+ C2
68 Cz
(02 + m+ n)
61 Cl
68 C8
+ (08 + « + p)
61
62
= ai
62 C2
68 Cz
— as
61 Ci
68 Cz
+ 08
61 Ci
62 C2
+ /
62 C2
68 C8
— m
61 Cl
68 C8
+ «
61 Ci
62 C2
\l
^2 C2
bz Cz
— n
61 Ci
68 Cz
'J
1 Cl
2 Cj
ai 61 ci
k 61 ci
Z 61 ci
=s
02 ^2 ^2
+
9n 62 C2
+
n 62 C2
a
8 68 <
58
8 6
B
Cz
P
68
cl
Cl
C2
862
ALGEBRA.
[§655.
9. The value of a determinant is not altered by adding to or
subtracting from the elements of any column the corresponding
elements of the other columns multiplied by the same factor.
For
052 + «i&2 + WC2 6a
Cl
a\hici
6161C1
C161C1
ca
=
(Za^aCa
+ w
&262C2
+ n
C262C2
C8
azhzCz
bzhzCz
Cs^sCs
But by Property 5 the lajst two determinants vanish, and hence
fli + fiibi + nci
61
Cl
Ol
61
Cl
02 + m62 + wca
&a
ca
=
02
62
Ca
as + wfes + ncs
6»
C8
08
68
C8
In like manner it may be shown that the value of a determinant is not
altered by subtracting elements as stated in Property 9.
This property holds true for a determinant of any order.
10. If the signs of all the dements of any column or row be
changed, the sign of the determinant will be changed.
For this changes the signs of each term of the determinant.
655. The application of one or more of the above properties
will make apparent the changes in the following determinants :
8
3
4
1
5
9
6
7
2
15 1 6
1 1 6
=
15 5 7
15 9 2
= 15
15 7
19 2
= 15
1
1
6
4
1
8
4
4
1
= 15
8
4
= 16(_ 16  8) =  15 X 24 = 360.
The second determinant is obtained from the first by adding the
second and third columns to the first, agreeable to Property 9. The third
determinant is obtained from the second according to 4. The fourth is
obtained from the third by subtracting the first row from the second and
third rows respectively, according to 9. The fifth is obtained from the
fourth according to 7. Then the determinant of the second order is
developed.
Trace the following changes in the same determinant :
8 16
3 5 7
r=
4 9 2
15 15
15
1 1 1
1
2
6
4
2
3 5
4 9
7
2
= 15
3 5 7
4 9 2
= 15
3 2
4 5
4
2
= 15
= X6( 4  20) =  15 X 24 = 300.
§ 656.]
DETERMINANTS.
858
Trace also the following changes :
8 16
3 6 7
=
4 9 ,2
8 1 6
8 16
15 16 16
= 16
1 1 1
= 16
4 9 2
4 9 2
7
1
1
6 9
=15
7
6
6
7
= 16( 49 + 26) =  15 X 24 =  360.
The foregoing determinant is a magic square, and can be reduced in
many different ways.
Trace the following changes in the magic square determinant
of the fifth order :
17 24
1
4
I 8 16
1
1 1
1 1
1
23 6 7 14 16
23 5 7 14 16
23 18 16 9 7
4 6 13 20 22
=66
4 6 13 20 22
=6.
5 4 2 9 16 18
10 12 19 21 3
10 12 19 21 3
10 2 9 11 7
11 18 26 2 9
11 18 26 2 9
11 7 14 _9 _2
18 16 9 7
65 90 70
65 90 70
= 66
2 9 16 18
2 9 11 7
= 66
6 26
2 9 11 7 '
= 66
6 25
35 95 46
7 14 _9 _2
1 _13 _42 19
113 _42 19
13 18 14
1 20 4
1 20 4
=126x65
1 6
= 125x65
15
=125x66
16
7 ]
L9
—
9
7 19
9
121 19
= 125 X 65
1 5
121 19
= 125 X 66(19 + 605) = 125 X 65 x 624 = 5070000.
The pupil may find amusement, as well as practice, in reducing this
determinant in a great many different ways.
656. By applying the principles now established, we can
readily solve groups of simultaneous linear equations contain
ing two, three, or more unknown numbers.
Take, for ezr«mple, the equations solved in § 647
flix + hiy {■ ciz = di,
a^x \ h^y + c^z = (^,
white's alo. — 23
G)
(2)
(3)
854
ALGEBRA.
[§ 657.
62 C2
63 Ca
and add the results, we have
If we multiply (1) by
(2) by 
61 Ci
63 C3
, and (3) by
61 ci
62 C2
+
62 C2
63 Cs
62 C2
^3 C3
a2
62
61 Ci
61 Ci
63 Cs
+ 08
+ &8
62 C2
62 C2
63 C8
C2
&1 Ci
63 Cs
+ C8
62 C2
63 C3
d2
63 Cs
+ (?3
62
ci\
C2 /
= di
But by Property 6 this may be written
bi ci
62 C2
ai
bi
Cl
a2
&2
C2
« +
as
bz
C3
61 &1 Ci
62 &2 C2
y +
63 63 C3
Cl
61
Cl
C2
62
C2
« =
C3
bs
C3
d\ 61 Cl
^2 &2 C2
ds bz Cs
Now, by Property 3 the coeflBcients of y and z vanish ; and hence
di 61 Cl
df2 62 C2
ds 63 Cs
a; =
«i 61 Cl
a2 62 C2
as &8 Cs
657. In like manner the values of y and z are obtained.
In finding these values, instead of rearranging the terms, v^e
change the signs of the determinants for y by interchanging
the columns containing the (Vs and 6's ; and for z, by inter
changing the columns containing the d's and c's.
Similarly, the values of the unknown numbers in a system of
linear equations with any number of unknown numbers may be
written down in determinant form.
§ 660.]
CUKVE TKACING.
S55
CHAPTER XXIII.
CURVE TRACING.
p'
668. The position of any place on the earth's Surface is
defined if its latitude and longitude are given, the latitude
being measured north and south from the equator, and the
longitude east and west from any convenient meridian.
North latitudes are defined as positive^ and south latitudes
as negative; east longitudes as positive^ and west longitudes
as negative (§ 60).
659. Similarly, we can locate the position of a point in a
plane if we know its distances
and directions from two fixed ^
intersecting straight lines. Let
these lines be drawn perpen
dicular to each other, as XX' xi
and YY, and let them inter
sect at O.
The line XX' is called the
axis of X or the axis of abscis
sas; the line YY, the a^s of
y or the axis of coordinates; and the two together, the axes of
coordinates. The point is called the origin. Each of the
four portions into which the two lines divide the plane of the
paper is called a quadrant, and the four quadrants are num
bered as indicated in the figure.
660. Distances measured along OX, or parallel to OX, from
YY to the right, are defined as positive; along OX', or parallel
to OX', from YY to the lefty as negaJtive. Distances measured
1^
8
O N
4 .
P
///
IF
Fig. 1.
356 ALGfiBKA. [§ 661.
along OF, or parallel to OF, from XX' upward, are defined as
positive; along OF, or parallel to OY^, downward, as negative.
Thus, if the point P is 3 units of length from YY^ in the posi
tive direction, and 2 units of length from XX' in the positive
direction (some convenient unit of length being taken), its
position is completely determined ; for from O we have only to
measure along OX a distance ON equal to 3, and then from
N measure upward and parallel to O F a distance NP equal
to 2, and we arrive at the position of the point P.
661. The two distances ON and NP are called the coordi
nates of the point P. Distances along the axis of x are usually
designated by the letter x, and distances along the axis of y by
the letter y. Then, in the example given for the coordinates
of P, we have a; = 3, and y = 2, visually written (3, 2), or in
general (x, y). The point (—3, 2) is found in the second
quadrant by measuring from O along OX' a distance ON
equal to — 3, and then from N measuring upward and parallel
to F a distance N'P' equal to 2, and we arrive at the position
of the point P'.
Similarly the point P", whose coordinates are (— 3, — 2), is
found in the third quadrant ; and the point P'", whose coordi
nates are (3, — 2), is found in the fourth quadrant.
662. It is thus seen, that, to determine the position of a point
in a plane, it is not sufficient to know simply the distances
of the point from the two lines XX' and YY', but the direc
tions must also be known. If the distances alone were known,
four points, one in each quadrant, would satisfy the conditions;
but, when the directions also are assigned, the position of the
point is completely determined.
Exercises.
1. What are the coordinates of the origin?
2. In what quadrants are the following points: (2, —6),
(3,5),(4, 1), (a,6)?
§663.] CURVE TRACING. 357
Draw the axes of coordinates and locate these points, using
J of an inch as a unit of measure ; and use the same unit of
measure in each of the following examples.
3. Locate the following points : (7, 8), (1, 0), (0, 3), (0, 0),
(5,  4), ( 1,  7), and ( 5, 0).
4. Locate the following points : (— 3, 2), (— 3, —2), (3, —2),
(0, 3), ( 1, 2), (1, 2i), (i,  2).
5. Draw the triangle which has for vertices the following
points : (1, 1), ( 3, 2), (0, 0).
6. Draw the quadrilateral having for its vertices the follow
ing points : (2,  3), ( 3, 4), (3,  4), ( 2, 3).
7. Draw the polygon having for its vertices the following
points: (2, 0), (1, 1), ( 1, 1), (0,  2), ( 1,  1), (1,  1).
The student is recommended either to provide himself with paper
ruled in small squares, or to draw on the paper two straight lines perpen
dicular to each other, and lay off distances from the origin along these
lines equal to ^ of an inch, and draw lines parallel to the original lines
from these points, thus dividing the paper into small squares.
GEOMETRICAL REPRESENTATION OF EQUATIONS WITH
TWO UNKNOWN QUANTITIES, X AND F.
I. Equations of the first degree in x and/.
663. Let 2x\y — l=:Ohe an equation of the first degree
in X and y. For every value of x we have a corresponding
value of 2^ ; so that, if x varies continuously, y also varies con
tinuously.
Let a number of corresponding values of x and y be found.
Thus,
If a; = 0,
y = l.
(2)
If «=l.
y= 3.
If a = 1,
y = l.
If a; = 2,
2^= 6.
If x = 2,
y = 3.
If «=3,
y= 7.
If a; = 3,
y = —6.
If a; =  4,
y= 9,
I£x = 4,
y=7.
If »=:6,
y = ll.
868
ALGEBRA.
[§ 664,
These values of x and y form the co5rdiiiates of a set of
points which can readily be located. When these points are
connected by a line, we have a geometrical representation or
picture of the equation. This line is called the locus of the
equation ; and hence the locus of an equation may be defined
as the line the coordinates of every point of which satisfy the
given equation.
If all possible values of x should be substituted in the equa
tion, and the corresponding values of y found, the sets of
values would represent the coordinates of an infinite number
of points ; and these points, if located exactly with reference
to the axes of coordinates, would form a continuous line or
locus. In practice, only those values of x are taken which
are most convenient, usually integral, and the coordinates of a
set of these points found.
664. To construct the locus of the equation 2ajfy — 1 = 0,
let us take the first four positive values of x given in § 663.
Draw the axes of coordination, and
assuming some convenient unit of meas
ure, say ^ of an inch, mark the paper
into squares, as in the figure.
The point (0, 1) is found at a, a dis
tance of 1 unit above O. The point
(1,1) is found by measuring a dis
tance equal to 1 on OX, and then a
distance equal to 1 below OX, parallel to
or, the point 6. The point (2, 3)
is at c, whose distance from O F is 2,
and from OX is — 3. The point (3, —6)
is at d, whose distance from OY is 3,
and from OX is — 6. Through the points a, 5, c, and d draw the indefi
nite line AB, and we have the geometrical representation or locus of the
equation 2a; + j/ — 1=0.
The foregoing process might be continued indefinitely for both posi
tive and negative values of x. Fractional values might also be used, and
points located between those found above.
\^ r
^
^
_5
 I K
y
X' «v x
^h
\
\
'S^
\
"^^
^
\
F' Ml
Fig. 2.
$685.] CURVE TRACING. 869
EXBRCISBS.
1. Construct the locua of 2x + y = 6.
ScQGBBTTOM. Computc & table of coire
aponding values of x and y, and, nith these
sets of corresponding valoea aa coordinates of
points, proceed as above. The line AB in
Fig. 3 is the locus of the equation.
2. Construct the locus of y = 5x+3.
The line Mlf in Fig. 3 is the required locus.
3. Construct the locus of
3x + 5y = 15.
Let X and y in turn ^nal 0, giving x = 6,
and y = S.
Locate the points (0, 3) and (6, 0), and fiu, 3.
through these points draw an indefinite line.
Construct the loci of the following equations :
4. 2a;+y = 6. 9. x — y = 0.
5. 7xiy = 0. 10. y = ix + 6.
6. iC + 5y = S. 11 2x^l=y.
7. y=2x + 2. 12 ^~2 ^ ^^^
8. xy = 3. 13. l + l = i
665. Each of the above loci will be found to be a straight
line, and it may be assumed that every equation of the first
degree in x and y is represented by a straight line; and, since
two straight lines can intersect in but one point, the point of
intersection of two straight lines is the geometrical solution of
the two equations of these lines.
Let us, for example, find the geometrical solution of the
<x + y + l = 0,
lx + 2y + i = 0.
860
ALGEBRA.
[§ 666.
Compute, as before, sets of values of x and y for each equar
tion, and construct loci, as in Fig. 4.
The lines will be found to inter
sect in the point P, whose coordi
nates are x = 2, and y = — 3.
These values of x and y must
satisfy both equations: for, since
every point in AB must satisfy
x\y+l=0, the coordinates of P
must satisfy this equation. Like
wise, since the coordinate of every
point in CD must satisfy
x + 2y + 4: = 0,
the coordinSites of P must satisfy
this equation. Hence «=2, and y= —3, satisfy both equations
simultaneously, and represent the solution of the equations.
14. Find the coordinates of the point of intersection of the
loci given in Fig. 3.
y
—
B
\
\
\
•V
n
\
\
i
"V
•*>
\
X
^
s.
\
X
"^
\
\
^
K
(2
s
)
p
'^
K
\
"V.
V
Q.
\
V
\^
Y.
f
^
Fig. 4.
Solve geometrically the following sets of equations :
(2a?f 32/ = 2,
(6a; + 3y = 4.
15.
16.
17.
I aj + 2^ = 4,
\x — y — (^,
ix + 2y = lS,
(3x + y = U.
18.
19.
20.
= 1,
(3a; + 42/ =
X9x2y=^ 4.
(4a;f 5t/ = 22,
\5x2y = ll.
II. Equations of the second degree in x and/.
666. The points whose coordinates satisfy the equation
y = x^ — 2x\2 do not lie in a straight line. They do, how
ever, lie in some line which is determined by the equation.
To construct this line it is necessary to find a number of points,
and then draw a continuous curve through them. Thus,
§ 666.]
CURVE TRACING.
861
For x = 0,
y = 2.
For x = l,
y=l.
For x = 2,
y = 2.
For a; = 2,
y = 3i.
For a; = 3,
y = 6.
For X = 3i,
y = 7i.
For a =  J,
y = 3.
For a; = — 1,
y = 6.
Y
—
\
/
\
/
/
\
/
\
/
\
/
\
y
X
X
r
Fig. 5.
The points tabulated above are plotted in Fig. 5, and a con
tinuous curve drawn through them. This curve is the locus
of the equation y = ar^ — 2a? + 2.
21. Construct the loci of y^ = 4aj + 2 (Fig. 6), and xy = 4:
(Fig. 7).
Writing the second equation
in the form y = , and assign
X
Ing both positive and negative
values to a, we get the follow
ing table of values for x and y.
For a; = 0,
y = 00.
For x = ±l,
y = ±4.
For x = ±2,
y = ±2.
For a; = ± 3,
y = ±t
^ 3
For a; = ± 4,
* * . •
Y
1*
V
^
^
,^
/
^'
J
/
X
(
X
\
\
s.
\
\,
\
'^..
^•5;
^..
^
r
For » = 0,
y = oo<
Fie. &
862
ALGEBRA.
[§667.
"" Y
T
i
n
^
^
^^
X\
^""^^
.^x^
X
1 '"'^1
4 K
>
" "4'T
r
"::.:.._... t
1
Y^
Fig. 7.
Construct the following loci ;
22. 4aj = 2^.
23. 3aj24y2=12.
24. 3aj2 + 42/2 = 12.
Plotting the points that corre
spond to the positive values of x
and y, we have a branch of the
curve in the first quadrant.
Corresponding to the negative
values, we have a branch of the
curve in the third quadrant.
It is thus seen that in this case
the curve is composed of two
branches, as indicated in Fig. 7.
26. aj2 + 2^ = 4.
26. a^ = 4:y,
27. a; + y + »* = — 1.
GEOMETRICAL REPRESENTATION OF THE ROOTS OF
AN EQUATION.
667. Let t/ = a^+2a; — 4. Compute
a table of values of x and y and con
struct the locus (Fig. 8). At the
points Pi and Pg? where the curve cuts
the axis of a?, we have y=0, and hence
ic^42aj — 4 = 0. The values of x for
these points are OPi and OP2; and as
these are the values of x which make
the expression a^f2a; — 4 = 0, they
represent the values of the roots of
the equation. These are readily seen
to be — 1 + V5 and — 1 —VS.
668. If the value of the absolute
term be increased, the curve will be
I
Y
\
\
/
\
\
1
\l
 ■
1
i
2\

■
—
J
x;
1
X
(,
i
—
\
\
t
\
/
r
Fio. 8.
§689.] CURVE TRACING. 36»
moved upward, and the points P, and Pj will approach each
other. When we have y = a^+ 2 a? + 1, the points Pi and Pj
coincide, and the curve simply touches the axis of x, or, we
might say, it cuts the axis of x in coincident points. For these
points we have a5* + 2a? + l = 0, the roots of which are readily
seen to be each equal to unity. This is the geometrical repre
sentation of the condition of equal roots. Thus, when the
curve touches the axis of a?, but does not cross it, the two values
of X are equal.
669. If the value of the absolute term be still further in
creased, the curve is moved upward still farther, and does not
cut the axis of x at all, or it is said to cut the axis in imaginary
points. In this case the roots are imaginary. Thus, if in ^ =
0^ + 2 a; + 2, we put y=^0, and find the corresponding values
of X, we get X = — 1 + V— 1 and — 1 — V— 1. These values
of X aye imaginary, and the corresponding points cannot be
located by means of the system of coordinates we are using.
HencCj^ when the locus does not cut the axis of Xy the roots of
the equation resulting by putting y = are imaginary.
Construct the loci of the following equations, and determine
the values of x which make 2^ = :
1.
y = iC^  1.
6.
y = oi^ — 2x.
2.
y = x^\'l.
7.
y = 2ix^ + i.
3.
4y = ^2aj2 + 16.
8.
y = ± 2 Va? h 4.
4.
y^ = 2x^.
9.
f = 16a?.
5.
43^ = a^.
10.
^ =z= a^ — 6 a?^ + 11 a? — 6.
864 ALGEBRA. [§ 670
CHAPTER XXIV.
PERMUTATIONS AND COMBINATIONS.
PERMUTATIONS.
670. The different orders in which a set of things can be
arranged are called permutations.
Thus, the permutations of the three letters a, 5, c, taken
two at a time, are
ab, hay ac, ca, be, cb.
In like manner the three digits 1, 2, 3, taken two at a time,
express the six numbers
12, 21, 13, 31, 23, 32.
671. The number of permutations of n things taken m at a
time is equal to the eontinued product of the m successive integers
from n to n — (m — 1).
For suppose the n things to be n letters, a, b, c, €?,•••, and
denote by "pi, "pj? "Ps? ••• "^Pn^ the number of permutations that
can be formed from n things taken 1, 2, 3, ••• m at a time.
The number of permutations of n letters taken one at a time
is evidently equal to the number of letters ; that is,
"p, = n.
If we put each of the n letters before each of the remaining
n — 1 letters, we obtain
"i^a = "Pi X (n — 1) = 71 (n — 1).
Again, if we put each of these **p2 permutations before each of
the remaining n — 2 letters^ we obtain
•i>3 = *i>8 X (n — 2) = n (n — l)(n — 2).
§ 674.] PERMUTATIONS. 366
Again, if we put each of these "pg permutations before each of
the remaining w — 3 letters, we obtain
"1>4 = "Ps X (n  3) = 7i (n  l)(n  2)(n  3).
We may proceed in this way until we obtain "p«_i; and, if
we put each of these "p«_i permutations before each of the
remaining n — (m — 1) letters, we shall obtain
"P«="P»iX[n(ml)]=w(nl)(n2)(n3)...[7i(ml)].
672. If all of the n letters are to be arranged together, then
m in the above formula becomes equal to w, and w — (m — 1) =
71 — w f 1 = 1, and
»p„ = w(n  l)(?i  2) ... 3, 2, 1.
Hence the number of permutations of n things taken all
together is equoU to the continued product of all the n integers
from n doum to 1 or from 1 to n.
Thus the number of permutations of four digits taken
together is 4 . 3 • 2 . 1 = 24.
673. The continued product of the successive numbers from
n to 1 or from 1 to n is denoted by the symbol [w, which is
read " factorial ti." Thus [5^ denotes 5.4.3.2.1.
The symbol n! is also used to express factorial n.
674. In the above formula for "p„, all the letters were sup
posed to be different. But if we suppose p of them, for exam
ple, to be the same, then the \p permutations, arising from the
presence of these letters when different, reduce to 1. If we
suppose another group of q letters to be the same, then \q
permutations, in like manner, reduce to 1.
Hence, to find the number of permutations of n things taken
all together, one set of p of them being alike and another set
of q of them being alike, we must divide ^p^ by \p and \q;
and the required number of permutations will then be expressed
by 1 , ; and similarly if there were three or more sets of like
things.
366 ALGEBRA. [§ 674.
EXERCISBS.
1. How many numbers can be expressed by 9 digits taken
4 at a time ?
Here n = 9, and m = 4, and hence
9p4 = 98 7 .6 = 3024.
The subscript 4 shows how many factors are to be taken, the super
script 9 being the first.
2. In how many different orders can 6 boys sit on a bench ?
% = 6. 5. 4. 3. 2.1 = 720.
3. How many permutations can be formed of the letters in
the word possessions, taken all together ?
Here n = 11, and the letter s occurs 5 times, and o twice.
"»« Hi 11 . 10 . 9 ... 3 . 2 • 1
»«"''« [^ = (Ig= 6.4.3.2. 1.2.1 = ^««^^
4. In how many ways can the letters a, 6, c, d, e, be
arranged, taken 3 at a time ? Taken 4 at a time ?
5. In how many ways could the seven prismatic colors be
arranged, taken all together ?
6. How many signals could be made with 5 flags of differ
ent colors, taken 1, 2, 3, 4, and 5 at a time ?
7. In how many ways could a party of 5 persons be seated
at a table ?
8. How many permutations could be formed from the let
ters of the word Columbia, taken all together ? Taken 5 at a
time ?
9. How many signals in all could be made with 7 flags of
different colors ?
10. How many permutations can be formed from the letters
of the word consonant, taken together ? From the letters of
the word Mississippi f
§ 678.] COMBINATIONS. 867
COMBINATIONS.
675. The different sets that can be formed from a given
number of things without regard to the order in which they
occur, are called combinations.
Thus, the possible combinations of the letters a, &, c, taken
two at a time, are db, ac, be, since db and ba, though different
permutations, are the same combination, as are also ac, ca, and
be, cb.
676. Any combination of m things will produce [m permu
tations, for the set of m things that form a given combination
can be permuted in [m different ways.
677. The number of combinations of n things taken m at a
time is equal to the number of permutations of n things taken m
at a time, divided by [m.
For if we denote the number of possible combinations of n
things taken m at a time by "(7„, then, by § 674,
... ng np ,^^n(nl)(n^2)..[n~(ml)1
" — [m
678. Since for every combination of m things which we take
out of n things we leave a combination of n — m things, it
follows that the number of combinations of n things taken m at
a time is equal to the number of combinations taken n — m at a
time ; that is, "Cm = °Cn_ni.
Exercises.
1. How many combinations can be formed with the letters
a, b, c, d, e, taken 3 at a time ?
Here n = 6, and m = 3 ; and hence
868 ALGEBRA. [§ 678.
The number of combinations of 6 letters taken 2 at a time is also 10,
since ^^ = 10.
1x2
2. How many different pickets of 5 men and an officer can
be made from a squad of 20 men and 3 officers ?
Here n = 20, and m = 6 ; and hence
^G, =20.19.18.17.16 ^ ^^^^
1.2.3.4.6
Since one of 3 officers may go with each picket, the whole number of
pickets is 16504 x 3 = 46612.
3. How many combinations can be formed of the letters of
the word longitude, taken 4 at a time ? How many combina
tions taken 5 at a time ?
4. How many sums can be formed of the digital numbers
1, 2, 3, 4, 5, 6, 7, 8, 9, taken 3 at a time ? How many taken
6 at a time ?
5. A number is the product of 5 different prime factors,
including unity. How many divisors has the number ?
6. How many different selections of 4 pieces each can be
made from 12 different pieces of money ?
7. How many different sums of money can be formed of
a cent, a threecent piece, a half dime, a dime, a quarter, and
a half dollar ?
8. In a school of 18 boys and 15 girls, how many classes
could be formed, each to consist of 4 boys and 3 girls ?
9. Out of the 7 prismatic colors how many combinations
of 3 colors each can be made ?
10. Company A contains one man more than Company B,
and the number of combinations of 3 men each that can be
made from Company A is to the number of similar combina
tions that can be made from Company B as 21 to 20. How
many men in each company ?
APPENDIX.
•o^
679. The following supplementary exercises are designed
for advanced classes, and especially for students who are pre
paring to enter higher institutions which may have exceptional
entrance requirements in algebra. Many of the problems have
been selected from recent college examination papers; and
each series contains as difficult problems as any found in the
papers examined.
680. Complex Fractions. (§ 273.)
Simplify
1 — a^ 1 — X
1.
2.
3.
1+a^ 1 + a?
1—05^ 1 —X
1 — OJ^/ X
l\y\l^x
1
11
1 a^^y^—x{y"
1
241
X
X
1
f
'A
1
2
X
white's alg. — 24
4.
5.
6.
7.
369
faJ^ + ll
x' + l
. 0311 >
1
a\
[l^x(xl)l
' + 0^
"l
y a?
if
*(
x + y I iB« + .v«
x — y Qi^ — 'tf
g; — y a^ — y^
x + y a^ + f
370 ALGEBRA. [§681.
a + b \ a + bj (3^f)(x+y)
1 x + y 1 a; — y ar'+(& + c)!g+6c a^ — cP
^' 1 y + x '^1 y — x ' ' iii? + {b + d)x + bd
y 1^ + 0? y y' + a? ti?+(a — c)x — ac
681. RaUonaUzing the Divisor. (§ 398.)
J VlO ^ (3+V3)(3+Vg)(v^2)
V5+V2V6' ' (5V5)(l+V3)
2. 4+V2V3 g __1
* 3V2+V3 l+V2+V^3
3. 1+V3+V6 . g V3 2V^2
1+V2V3 • 2_V3 2+V=^'
4 V(3 + 2V2)V2 J
• V(3  2 V2) + V2' 10. ^25^20 + 2^^
g^ ^ + V 6 ^ SuGGBSTiON. — Multiply both
5V32V2V32+V50* terms by \/6 + v^.
. 3 + 2V23V32V6 , 1
6. — • 11.
1 _. V2  V3  V6 V25+ VlO +V4
12. (Jl±^^^l^\J ^'^' .
\\ab yia\bJ\ (a\by  ab
682. Equations involving Radicals. (§ 410.)
Solve
1. Va4V4a+ic=2V6fa;. 3. Va—x\\/b—x=
^b^
X
2. ^ 4, ^ ^ J^ . 4. V2+a;+^fa?4V2a;
' Vicl Vic+l Va^1 =V2ic.
§ 684.]
APPENDIX.
371
5.
3*
6.
V5a + 2 3V5^8
7.
8.
9.
10.
Suggestion. — Equations 5 to 10 may be solved by proportion.
11.
Va + Vci — X 1
Va — Va — X ^
3^+ vr^_a+ VF^^
12.
\/x\a\Vx—a _ a\b 13.
Vif+a— Va;— a a— 6
V4a;41 +V4^ _^j
V4iB + l— V4« 14.
Vl4a;+Vl4 + a; = 4
f •y/2x — y = Vx — y + 1,
■y/x y/y
' +4 = 4.
VS V3/
1 _1
x" — 6a5 " = 1.
683. Imaginary Numbers. (§ 420.) Simplify
1. (65V^)(5 +3V^). 5.  8 V26 5 2 V^:^.
2. (2.fV^r6)(3 42V^=TI)). 6. (3V^l
3. v^=nroTV6.
4. Vl55V^^.
9.
7. (3V5)
8. (3Vl)5(2 + V
(1+V^=^+V^(1+V^V
1).
12.
3).
10. (9V^iiV^^)^(V^+V^=^).
11. (2V6V^H(V^=^V2+V^^).
14. V40  42 V^.
15. V5 + 12V^=l.
a^b^Z—l a — bV^^
a — bV^^ a + ftV— 1
13. V34  12 V^. (§407.) 16. V_26 46V^=^.
684. Fractional and Negative Exponents. (§ 436.)
Extract the square root of
1. a^y"^ — 4 x'^y'^ + 6 — 4 x'^y^ + x'^y^.
2. a^ — a*VS + 2a*V^ + 4ic(aj — fa).
Extract the cube root of
3. 8 «*  12 aj"'V 4 6 ic"*  a?"*.
4. ai9a* + 33a^63a + 66a*36a* + 8a*.
872
ALGEBRA.
[§ 685.
Resolve into factors
6. x — y. 7. a"^ — &.
6. aj*"+^ — x~l
Simplify
8. x^ — y\
9. x — x^y^{\y,
3 3 X
10. a;"2 2a;"^2/*+y.
11.
12.
J
a^a ^.
18. ^^^ r X
aj^y
aJ^2^
»
a
3e+jf\ X
a" \'*
a?
a»
' — X
y otyiwab
13. [(a;^"/*)^ «*]«•
19. a^6 ^ X a h ^ x a^6'
20. (a*6~^)* X (a«6~^)i
2/. 2
14.
a26
X
a'hb'
3a
3a
a^ + b ''" a '
21. x^y ^XX "225* x(2/22J*)^
( _i 1 1 i^V"^'\5
15. \a ^x^yax 3 Vic^y .
9"^* X ^3"+^
ah 3a
ya5\ 2 n 2J
16.
17.
3V3"
92 ^ 16*
81"'^ ^ 4*
26.
e
4e
i\/r
+
22. [(x^y ^^yy.
23. (a;^»*^* ^ aj^"+'>*) x a?*«».
1
24. [(«"•")'"+'* X (a"*)""]*.
25. (e' + e*)*(e'0'
rV^1
xver>«
2 ; • V 2V1
27. ic^+^ft^^' X afY"+^ ! m^+^y^,
28. f (»  1)* (aj + 1)"^ + i (^ 1)^ («  1)"*.
29. (a6^ + l)[ai(l+V^3)6]x[ai(lV^3)6].
685. Quadratic Equations. (§§475485.)
Solve the following groups of equations :
J [0^ + 2/^ = 189,
V2/^ + 2/' = 10,
icy* + 2/ = 4.
(x' + xy\4:f = e,
' [Sx' + Sf^U.
(x'+fl5{xhy)=My
*• l2a:2/431(aj + 2/) = 289.
2.
5.
6.
X" + 2/** = ct,
xy=b,
ra^ + 2/* = 17,
^ a? + y = a?2^ + 1.
§ 688.] APPENDIX. 873
686. Equations of Higher Degrees. (§§ 471, 472.)
1. 3aj8 + 7ar'h3aj + 2 = 0.
Suggestion. — Divide by x + 2, and then solve the resulting quadratic.
2. 64a^39iB*26aj416 = 0.
3. i»* + 6aj3 + 5ic2 12a;12 = 0.
4. afi + 15a3^jUa^ = 0.
5. aj*10aj8 + 35iB250a; + 24 = 0.
6. aj*a?*64a5* + 64a? = 0.
7. a^2aj8 40.132 = 0.
687. Proportion and Progression.
1. If a:b = c:d, prove a:b =\^3a? \ 5c^:VSb^ ^ 5d\
2. Find a mean proportional between (1) 4a^ — 3« — 1
and xl; (2) 2aj3 and 2a^4aJ^4aj 3.
3. Two numbers whose difference is d, are to each other as
a to b. What are the numbers ?
4. Prove that in a geometrical progression s = ^ ~^ '
?• — 1
5. Show that if a f 6, 26, 6 + c are in harmonic progres
sion, then a, 6, c will be in geometrical progression.
6. The sum of the first six terms of a geometrical pro
gression is 728, and the sum of the third and fourth terms
is 72. Write the series.
688. Convergent Infinite Series. (§ 573.)
7. Show that in a decreasing geometrical series to infinity
a
s =
1r
Find the sum to infinity of
«• 81^. 12. 1+A. + U....
10 1 1 i  _^_
874
ALGEBRA.
[§ 689.
Find the value of the following recurring decimals :
14. .24. 15. .851. 16. 62l. 17. .037. 18. .12135.
19. Show that 0^a=0; a50=oo; and 050= any number.
689. Exponential Equations. (§ 613.)
Solve the following exponential equations :
1. 23'+« = 5*^ 3. 7*' = 2»+*.3*.
2. 3^^ = 2.5'+^
4. ^1""+* = %.
5. 3**^' = 4^
6. 2*"i = 3^».
690. Undetennined Coefficients.
Resolve into partial fractions
1.
2.
3.
(x  S){x h 1)*
3a^^_3a; + 18
«»8
Expand into series
1x
4.
5.
6.
lH2a;
a^(x  l)(x + 2y
2a^H^l .
2aj*f a;3*
7.
1 fajhoj*
8.
1f ajho*
> .
x — a^
9.
l42a^a^
691 . Determinants .
Solve the following linear equations :
1.
2.
3.
4.
5.
6.
a;2y = 3,
2x{5y = 5.
r3aj42/ = 4,
6x\6y = l,
6x — 5y = 7,
7a; 22/ = 12.
Expand the following determinants :
'8aj + iy = 26,
4a? — 5y = — 8.
Sx + 5y = a,
4aj42y = &.
aa; h &y = c>
ca; } dy = a.
7.
a;f 2/ a;— 2/
8.
aJh2/ ar^+2/^
05— y «^— y*
9.
a;2/ x^—xy\y^
ANSWERS.
Note. — When several examples are given under the same number, in
most cases the answer to the last only is given. Answerswhich, ii' given,
would destroy the utility of examples, are omitted.
Page 11.— 2. Son, $14; father, $42. 3. Vest, $4; coat, $20.
4. Less, 7 ; greater, 35. 5. Less, 9 ; greater, 27. 6. Shorter, 5 yds.; longer,
25 yds.
Page 12. — 7. B, 50 sheep ; A, 100 sheep. 8. Less, 18 ; greater, 72.
9. 20. 11. Father's age, 46 yrs.; son's, 15 yrs. 12. 135. 13. Saddle,
$6; harness, $18.
Page 13. — 16. Shorter, 45 ft. ; longer, 76 ft. 16. 50 and 70. 17. 1st
person's, $576; 2d person's, $1226. 18. Chain, $20; watch, $66.
19. 1st party, 158 votes; 2d party, 206 votes. 20. Men, 47; women,
62 ; children, 218. 21. A, $3000 ; B, $ 1500 ; C, $7500. 22. 1st, 7 yds. ;
2d, 21 yds.; 3d, 14 yds. 23. Shorter, 15 ft.; longer, 30 ft. 24. A's
share, $1200; B's, $2400; C's, $4800. 25. A's share, $4; B's, $2;
C's, $ 12. 26. First, $400 ; second, $600 ; third, $800.
Page 14. — 27. A's age, 40 yrs. ; B's, 20 yrs. ; C's, 35 yrs. 28. 16.
29. One, $60; other, $80. 30. B's age, 14 yrs. ; A's, 56 yrs. 31. Boys,
80 ; girls, 160. 32. Buggy, $ 60 ; horse, $ 120. 33. Mother's age, 46 yrs. ;
daughter's, 15 yrs. 34. A's share, $86; B's, $40. 36. 30 and 46.
86. Smaller, 79 acres ; larger, 121 acres. 37. 186 sheep. 38. Standing,
35 ft. 'y broken off, 56 ft. 39. B, $2000; A, $3500.
Page 15. — 40. Each son, $3675 ; widow, $ 14,700. 41. Shorter, 24 in.;
longer, 40 in. 42. 18. 43. 1st, 12 ; 2d, 24 ; 3d, 40. 44. Father's age,
48 yrs. ; son's, 24 yrs. 45. Mother's age, 36 yrs. ; daughter's, 16 yrs.
46. A's age, 72 yrs. ; B's. 36 yrs. ; C's, 18 yrs. 47. Father, $80 ; elder son,
$ 40 ; younger, $ 20. 48. 1st candidate, 137 votes ; 2d, 97 votes ; 3d, 72 votes.
Page24. — 1. 16. 2.18. 3.6. 4.31. 5.18. 6.26. 7.72. 8.20.
Page 25.9. 9. 10. 51. 11. 31. 12. 28. 13. y. 14. 2x^.
15. 2 y, 16. 4. 17.  a. 18. ab ^ h. 19. 64, 20. 4 a + a6 + 2 6.
21.2 6. 22. 2a + 7 6. 23. 2bK 24. b^. 25. 2xy. 26. a;y2y«.
Page26. — 1. 10. 2.26. 3.7. 4.24. 5.51.
Page 27. —6. 6 a. 7. 6 x. 8. 64 x  18. 9. ^xy  13.
Page 29. — 1. 23. 2. 56. 3. 4. 4. 45. 5. 21. 6. 4. 7. 38.
8. 0. 9.  1. 10. 88. 11. 37. 12. 14. 18. 62. 14. 18. 15. 28.
375
376 ANSWERS.
16. 35. 17. 92. 16. 0. 19. 7. 90. 22. 81. 11. 98. 169. 83. 28.
94. 188. 85. 7.
Page 39. — 1. z = 3. 8. z = 3. 8. z = G. 4. z = 8. 6. z = 8.
6. z = 7. 7. z = 4. 9. z = 2. 9. z = 6. 10. z = 3.
Page 40.— 11. z = 2. 12. z = 7. 13. z = 6. 14. z = 5.
15. z = 11. 16. z = 7. 17. z = 3. 18. z = 10. 19. ^ = 10. 20. z = 3.
81. Com, 120 bu. ; wheat, 240 bu. ; oats, 360 bu. 82. 1st, $ 14 ; 2d,
$28 ; 3d, $84. 83. Quarters, 4 ; dimes, 16 ; nickels, 32. 84. 1st year,
6900; 2d year, $1800; 3d year, $3600. 85. B*s, 62000; A*s, 64000;
C's, 6 6000. 86. 9 and 27. 87. A*s, 6 28 ; B's, 6 7. 28. Younger, 8 y rs. ;
elder, 24 yrs. ; father, 48 yrs.
Page 41. — SO. In quarters, 63; in dimes, 6120; in nickels, 660.
81. Younger, 6 18 ; elder, 636 ; father, 6 108. 38. A, 30 yrs. ; B, 40 yrs. ;
C, 60 yrs. 38. 1st farm, 100 acres ; 2 J farm, 80 acres. 34. Each sister,
170 acres; brother, 300 acres. 35. A, 6^0; B, 6130. 36. B, 62420;
A, 64840 ; C, 6^760. 37. 9 hours; one travels 27 miles, other 36 miles.
Page 42. — 38. First, 50; second, 80; third, 65. 39. Serge, 75^ per
yiurd; silk, 61.50. 40. Oranges, 4 doz. ; lemons, 8 doz. ; pears, 16 doz.
41. Vest, 65; trousers, 6 10; coat, 615. 48. Bonnet, 65; dress, 613;
eloak, 617. 43. First, 625; second, 640; third, 635. 44. 61.
45. B, 40 acres; A, 60 acres ; C, 60 acres. 46. 9 hours.
Page 44.17. 15(a + 6). 18. 13 zy(a  6). 19. S2Vab.
80.  llmnVx'.
Page 45.— 87. 5(a + c). 88. 0. 29. Vx. 30. baVxy,
Page 46.38. Aa'^h\\bah\2ah\ 39. 5ax+2az2zV 40. Oz'y.
41. z« + 3z2y 4 3zy2  yS. 42. %anx'^  3nz8. 43. 8a6 + 46 + 3c.
44. 7a6 + 56c3a6c. 45. 3z2y + 6zy".
Page 47.8. 16a. 3. 3a + 6 + c. 4. lly22r. 5. z« + 42«.
6. 7a2+96y+27. 1. Aay'^+cz^+lO. 8. \1 ahd^+Zb^y^, 9. QoC^bc
•^ 22 ab^c  7 abc^ 10. 2 a^ 2 62+ 7 d^. 11. 24z2+ 12y2. 12. 3pzn.
13. _5y2+10l2;2. 14. 3a6+85 6c+2 ac. 15. 14z2+75y2_342.2.
Page 48. — 16. 23 z2+ 25pxy  5 y2. 17. 5 aa^  4 bz^ + 2 cz. 18. 7.
19. wn(z2 z + y); 3a(z«z^zy32j^). 80. z(z  a  64c)+ a6 ;
a%b  c)  62(d  c). 81. z2(z  3 y + 5j^2) ; 5(x2  2 y) + 6(z + 2 y^).
22. (ac)x\(bd)y; a(VxVy{V2z).
Page 50. — 10. l2xySz^{Sy. 11. 4zy. 18. 26. 13. 4a«4rt6.
14. 2a + 2 62c. 15. 3a210a63c. 16. 5z2+2zy10. 17. 6z^y
+2y». 18. 6z2y2ys. 19. 2a2362+10. 80. 2z82az2f Sa^z.
21. 26 + 2d2e. 88. 2c. 28. 2pv, 24. a24.2a6+362. 85. 8/)%
 5pz2. 86. 9x8 + 31 ax^ 4 ISa*. 27. z*  4z8y 4 12zV + 2y*.
Page 51.— 28. 3(z24y2) + 10(a2_62). 2d. 5y/c^^T^^Vifi^\5y/S.
30. z8+z2y8y2. 31. Sxh/. 1. 9(z4y). 8. ZVx^. 3. (3a+6)
(z2 y2). 4. 16z2 (a  z^). 5. z  5Vxy + 3y. 6. 2a2f 3az + Sz*.
7. 5z2c2xc. 8. (ab)(x{y\z). 9. a(5z). 10. z2+4zyy».
11. 3az226x2. 12. 7x8y+x2y2+5xy8+7 y* + 8. 13. 22(a + 6)(a6).
Page52. — 14. 2(x2y2). 15. 22 (0^+ 62) + (a^ 6*). 16.22(a+6+c).
17. 2c. 18. 6xya + 2y«. 1. a  6. 2. a 2c. 3. 4a2+ac.
A>JSWERS. 377
Page 53.4. h^. 6. 4a;H5. 6. x^^xy. 7. x^+xy, 8. 17a;2.
9. a2+ 62_ 2 a6. lO. a:* y*. 11. 4 ax8 + 6a;2 4. feo; + 9. 12. 4 a2a;2 ^. ^xK
13. 2 a262_ 2 ab. 14. a. 15. a  6. 16. y. 17. 6 m  8 n  p. 18. 12 m
 8 ri  r. 19.  a  7 6 + 2 c. 20. 6x  11. 21. a + 4 y. 22.6.
23. a b\d, 24. a6+6c+c. 26. x. 26. 2+3 aa*. 27. a+6+7c.
28. 18.
Page 56.2. Sa^cSabcSac^. 8. 6x^y+6xy^bxyz. 4. 3rt*6
 a862 + 3a2&8. 6. 12 x*y2 _ g a;8y8 ^. 20 a;2y4 + 4 icyS. 6. 3ac2x2
2 c8x8 4 6 c^x*. 7. 4 a*6x  24 a^x^ _ 4 a^f^^* + 4 a^x&. 8. 6 a^y^ 
6a*x&y2_i5a&a;8y2+5<j2ca;22^. 9. Ga^b*x^42ab^x^y^6ab^3^. 10. 6x2/*
2a:2y8 + xV 11 ^a^bdiabcdiiacd^. 12.   a'6*  y^^ a^fts
+ fa868. 18. 2 a.22^ _ J x8y5 + J a;*2^  I a;y7.
Page 58. —8. 4 a2 4. 8 «» + 2 a*6 + 3 a*  5 ««&  10 a^b^  17 a«62 ^
150^6'*. 4. 4x6  9x* + 31 ajs _ 4a;2 + 27 jc + 16. 6. x^ + x2y _ 6x2/2.
6. 2 x8  x2  16x + 16. 7. x2 + 7xy + 122/2. g. 6m*  13 w2»  m2 
6n2 + 3n. 9. cfi  a^b\ 10. a^  3a25 + Saft^ _ 58. xi. ax28a2a;
+ 16 a*. 12. X*  2/*. 18. 64x*  9 a^y\ 14. x* + x22/  xy^  2^.
16. m*w2 + 2m8n8 + 3m2nm2n*3mw2. 16. x*  9x8+22 x^24x.
17. X* 2x22/2 +2/4. 18. 2x510x*2/+9x32/217x2/*+42/^ 19. 25x6a4
 30 x^a^ + 14 x4a«  8 x^a^ + 3 x2a8. 20. a^ + a*  3 a86 + 3 a^b'^  a^b^
 a*b  tf 68 + 06*. 21. x^x^y2 x^y'^ + 2 x^y^ + X2/*  2/^. 22. x8 +
3x2/ + 2/8_i. 23. a«3a629a*+ 101a^ + 102a2_644a + 480.
24. z^yz*2y'^z^\2y^z^jy*zy^. 26. x64xS+3x*+2x8+4x216.
Page 59.1. x8+l. 2. 1x*. 3. x3+ 3x22/ +3x2/2 +2/8. 4. a^Sa^b
+ 3a62 _ 68. 6. x*  2x22/2 + y*. 6. a* + 4a2&2 ^ 1654. 7. i5a;4 4.
16x82/  17 x22/2 + 8x2/82/*. 8. 1  a2 4. ^3 _ 2 a* + «&. 9. 26x6  x*
 2 x8  8 x2. 10. X*  2/*. 11. a8 _ j)6, 12. 3 a*  26 aS^ + 37 ^252 _
14a68 + 3o2 _ 6 aft 4. 2 62. 18. €fia^^a*ba*b^+a^b*ab*\b^b\
14. x5 + 6x*a + 10x8^2 4. I0x2a8 4. s^a* + a^ 16. x^  x^ + x2  1.
16. 36x*39x8+13x226x+4. 17. x*+9x2+81. 18. x541x120.
19. 20x  56x2 + 10x8  32x* + x^. 20. 3x8 + Sx^y* + Sy^ 21. a^ 
3 abc + 68 + c8. 22. a*  2 rt252 4. 4 ^5^2 + 6*  c*.
Page 60. —23. ofiy^. 24. x*a*. 25. 6*562c2+4c*. 26. a262;
X22/2. 27. 4a2_52; 4x292/2. 28. w2_^2. 4w2n2. 29. x8 
3x24x+12. 80. x*2x2+3x2. 81. x*+2ax89a2a;2_2a8a;4.8a4.
32. a*+2a86_2a686*. 33. x*a2x22ax2+2a8. 34. a22a6 + 62.
35. x8  3 x22/ + 3 X2/2  y^. 36. 8 x8 + 12 cx2 + 6 c2x + c8. 37. a*  4 a^b
+ 6a262  4a68 + 6*. 38. a*  4 a86 + 6a262 _ 4^68 + 6*. 39. x2  2xy
2xz + 2/2 + 2yz + ^2. 40. 27 x8 + b\x^y + 36x^2 + 8 2/8. 41. 16x* 
96 X82/ + 216 x22/2  216 xy^ + 81 2/*. 42. 3 d^ + a6 + 6 ac. 43. a8 _ 2 0525
4a6268. 44. 2a6 + rtc. 46. 3a6. 46. a2  4a6 3 62.
Page 61. —47. 6 a2 4. 2. 48. x* + 3x22/2 + 2/*. 49. 4 X82/  4 X2/8.
60. a8+4a2+ 6a + 2. 61. 0.
Page 63.6. 2 6*c. 6. 9a2y2. 7. 6m*wx2. 8. 2ac. 9. 8g.
10. 40a8x22:. 11.7 m?/. 12. 4a*6. 13. 12x. 14. 8a2a;8y. 16. 7 nr^.
Page 64.2. x^^xy 2/2. 3.  a6 + 2 a62  3 62 + 6. 4. xy
2x22/2 32J. 5. x8 + 2«x22/2a2x2/2 + 2a82/8. 6. 2 2/  4 X82/2  x*.
7. 6x8+2 ax2 10 a2x. 8.  6 x + 4 2/  9 z. 9. 6 0^62 4 a68+ 3 a6c2.
10. 6x+ 14x2/2^ + 6x82/ 20 2/2.
378 ANSWERS.
Page 66. —6. 2 x«+ 4 x^+ 8a; + 16. 6. a^+ Sxhf + 33cy«+ s^. 7. a:H
ary + yK 8. x*4 y*. 9. 2a^Sab +465. 10. 6a26  bab^. 11. x + y.
ia.xy. 13.3a + 26. 14.6a;7y. 16.4a;+12. 16. a^ + a6 f 6«.
17. 1+ a: + x2. 18. \i x* + ^ x^y '\ 4 y^. 19. 25 m* 35 itiH 49. 20. m^
niH + mn^  n'. 21. a* + a^6 + a'^b^ + aft* + M. 22. x* + x^ ^. y4,
28. 25a:H10a;+l. 24. 36x4 24aa;24 4a2. 25. x^ 6x4 4. 26. x+a.
27. 9x* 16x2125. 28. a< a^b^\ b*. 29. 81x*4 27x^4 9x2+ 3x + 1.
80. 144m<+84w2+49. 81.6x23x49. 82.2x26. 38. x245xl.
Page 67.84. 1  y2. 35. a* 4 a;*y2+ «V+ V^ 86. x* x2y24 y*,
87. a«  a* 4 1. 88. 1  3 x 4 2 x*  x*. 89. a'^x* 4 ab^'^y 4 6*y2.
40. a 4 ft. 41.4x212x49. 42. 1 4 2y 4 3y2 4 2y3 4 y*. 48.3x2
xy2y2. 44. 4 43x42x2. 46. x*  6x811x2 499x4 10.
Page 68.— 1. j aft2 4 i ft^c 4 tV 2. 2a2 a6 4 Jft2. 8. ixVy
45xy. 4. ia24T^a&462. 6. \oi^^j\x  j\. 6. Ja^  024 ^i^a 4 i
7. J a 4 J ft. 8. 4x2  ix 4 T>5. 9. ix8  .^^x^  A^^ + tV 10. ^ a*
y\ aft2_ J 63. 11. xs^. 1 a;24 J X J. 12. i x  . 18. i a 4 i ft. 14. J x2
 j^x J. 16. Jx4 iy.
Page 69.5. a*  2 a2ft2 4 ft*. 6. 2 5x4 10x2  6x8 4 5x* 4 2xS.
7. 6 x6  26 x* 4 49x8 66x2 4 3.3 X 10. S, x^+x^y*+y\ 9. a^Qa^
4 13 a* 14 0584. 10a2_4a4 1. 10. a641o 120. 11. Sa^6a^
5a416a8_ i2a26a3. 12. xS4 i«* f i^a^ f Ji*^  ^x  t\.
Page 70. — 18. am3^—anx^\(bm—ar)x^\{cm—bn)3^+(dni — cn—br)x^
(dn^cr)xdr. 14. x66x53x4440x84 24x224x44. 15. a^Ox^
4 16x* 20x84 15x26x41. 19. x23xl. 20. 2x2 43x42.
Page 73. — 4. x2l. 5. x23xl. 6. X24 x 4 2. 7. 1 4 x 4 2 x2.
8. x8x23. 9. x245xl. 10. x843x2y43xy24y8. 11. x2JxJ*
Page 77. — 8. x = 10. 4. x = 5. 6. x = 6. 6. x = 3. 7. x = 6.
8. x = 4. 9. x = 3. 10. x = 6J. 11. x = 7. 12. x = 6. 18. x = 4.
14. X = 9.
Page 78. — 15. x = 1. 16. x = 10. 17. x = 7. 18. x = 9. 19. x = 4.
20. X = }. 21. X = 6. 22. X = 6. 28. x = 1. 24. x = 6. 25. x = 3.
26. X = 2. 27. X = 9. 28. x = 14. 29. x = 15. 80. x = 2. 81. x = 6.
82. X = ft. 83. X = 2. 34. X = 1. 85. x = — 1. 86. x = 4 a. 37. x = 3.
88. x = 6. 89. x = . 40. x = 2a2ft. 41. x = a 4 ft. 42. x = a4ft.
a2
43. X = 2 ft. 44. X = aft.
Page 81. — 15. 38 and 65. 16. 6. 17. 22 and 32. 18. Man's a<;e,
25 yrs. ; brother's. 30 yrs. 19. 15 yrs. 20. Son's age, 9 yrs. ; father's,
36 yrs. 21. B, $252 ; A, $324 ; C, $424.
Page 82.22. Horse, $120 ; buggy, $80. 28. Less, 27 ; greater, 46.
24. Boys, 78 ; girls, 65. 25. 15. 26. First, 4 ; second, 12 ; third, 24.
27. Unsuccessful candidate, 1030 ; successful, 1630. 28. A's share,
$1243; B's, $904. 29. 112 sheep. 30. A, $36; B, $40. 81. 60 pieces,
32. Horses, 4 ; cows, 12 ; sheep, 80. 33. Half dollars, 60 ; dimes, 40.
Page 83.34. 60 children. 35. A, $3500; B, $4000; C, $7600.
86. Eldest, 24 yrs. ; second, 21 yrs. ; third, 18 yrs. ; fourth, 15 yrs. ; young
est, 12 yrs. 37. Less, 7 ; greater, 11. 38. C's share, $440 ; B's, $1320;
A's, $2640. 39. Women, 18 ; men, 22 ; children, 50. 40. 6 vessels.
41. First, $ 1300 ; second, $ 1500 ; third, $ 1.300 ; fourth, $ 900. 42. 103 gals.
43. Sister's age, 6 yrs. ; brother's, 12 yrs
ANSWERS. 379
Page 84. — 44. Artillery, 260 men ; cavalry, 450 men ; infantry, 3800
men. 46. 10 hours. 46. 50 and 20. 47. $2400. 48. Left pocket, 15 ;
right, 20. 49. Youngest, $ 425 ; third, $ 475 ; second, $ 625 ; oldest, $ 575.
60. Quarters, 5 ; dimes, 16 ; fivecent pieces, 75. 61. 45 miles. 62. 5
miles.
Page 85.4. a^ + Oacf 9. 6. 25 + 10a;  x^, 6. 9a^ \6ab f b\
7. lQx^\Sxy + y'^. 8. 4 a^2 ^ 12 ax f 9. 9. 9 a^ + 30 a6 4 25 &2.
Page 86.— 18. 2510a;4x2. 14. a'^2a^l^\b^. 16. 9a^Qab\bK
16. a;2  10 xy + 25 y^. 17. a^^  8 ax + 16. 18. 16 a'^ _ 24 a6 + 9 62.
Page 87.19. a^ + 6ax + 9x^. 20* x^6xyh9y^, 21. 4m^ +
4w+l. 22. 14 771+4 w2. 28. a^^a'^bh^b^. 24. 4a212a&+962.
26. 9 a* 12a^62+4 6*. 26. a2x2  4 axy + 4 y2. 27. a^fea f 4 adc^ f 4 c*.
28. a«62  4 a^bcx + 4 c2a;2. 29. a^y^ + 50 a:*y + 625. 80. 16 a2ft2 _ iq ^bxy
+ 4 a;2y2. 83. 9 x2 _ 9. 84. 4 a2  9 62. 35. aH^  62. 36. 19 x*y2.
87. 16 a*x2  9 yK 88. 25 a*6*  1.
Page 88. —41. x"^ ^ y'^ ■{■ z^ \ 2xy  2xz  2 yz. 42. x'^ ^ y'^ + z^ 
2xy + 2xz2yz, AZ. x^ \ y^ \ I h 2xy  2x2y, 44. x2 4. yS __ ;j2
2xy 2xz\2yz. 46. 4 a2 + 62 + c2  4 a6 + 4 ac  2 6c. 46. a2 f
962 + c2 + 6 a6  2 ac  66c. 47. x2 + ^2 ^ ^3,2 ^ ^2 4. 2^2/ + 2 x;? + 2rx
+ 2yz {2vy \2vz, 48. x2 f y^ + ^2 4. ^2 _ 2xy 2xz 2vx +2yz\
2vy\2vz, 49. a2+4 62+c2+4cZ2_4a62ac+4ad+4 6c86<?4c(i.
60. a2 4 62 + c2 4 ^^ + 2 a6  2 ac  2 a(i  2 6c  2 6(i 4 2 cd. 61. a* +
6* 4 c* 4 <«* 4 2 rt262  2 aV.2  2 a^d^  2 62c2  2 62(f2 + 2 c2d2. 62. a* 4
64 4. 4c2 4. d2 4 2a262  4a2c + 2a2d 4 62c 42 62^4 cd.
Page 89. — 68. x2 + 7x 4 12. 64. x2  lOx 4 24. 66. x2 4 6x  24.
66. x«2x63. 67. x23x18. 68. x^8x65. 69. x216x460.
60. x2  3x  108. 61. x2  24 X 4 135. 62. x2 (a + 6)x + a6.
68. a2 4 5 a  QQ. 64. 81  9(x + y){ xy, 66. x2  (2 a + 3)x + 6 a,
66. x2 + 2 ex  15 c2. 67. x2  4(6 4 l)x 4 16 6. 68. x2 + a6x  2 a262.
Page 90. —69. 3x2  26x + 35. 70. 6x2  11 x  35. 71. 12 m^ 
31m4 20. 72.3x210x4 3. 78. 3x26x2. 74. 5x2 47 x + 84.
76. 6x2 xy 22/2. 76. 6  13 a6 + 6 a262. 77. 2 a2  a6 362.
78. 5x2 + 6 xj/  8 y2. 79. 2 w2 4 mH  3 m2n2. 80. 3 a2 + a6  2 62.
81. 3x266x262. 82. 10 2/2 _ 12 a?/ 4 2 a2. 88. 4x2 4 4x2/  3v2.
84. 15 a2 4 4 a6  35 62. 1. x  j/. 2. a2 4 6. 8. x  3. 4. 3a4*6.
6. 2ax+3. 6. bxy. 7. 3a2 6. 8. 2x232/2. 9. x+8. 10. a10.
11. X — 2a. 12. x3c.
Page 91. — 18. x + y, 14. x2  2/2. 16. x2 + 4. 16. 3 x  3.
17. 2rt+36. 18. 3ax46. 19. 10+3a6. 20. l\Sx% 21. 5ax2l.
22. 4 x^y 4 3.
Page 92. —28. x^  xy + y^. 24. x^ {• xy + y\ 26. 4 a2  2 a6 + 62.
26. 4 a2 h 2 a6 4 62. 27. x^ + 3 6x 4 9 62. 28. 1 4 2 2/ + 4 2/2. 29.4 2/2 4
2 2/ + 1. 80. x22/2 + xyz + z\ 81. x^y^  xyz 4 z^^ 82. 9  3 xy 4 x'^y^.
83. X* 4 «22/2 + y4. 84. X*  x^y^ 42/*. 88. 1 + 4 x?/ 4 16 x^y^. 86. 9 x;^
+ 6x2/ +4.
380 ANSWERS.
+ 2x2fa". 47. x*2a5*+4 0588x2+ 16 a;32. 48. 7?+Sx^h9x+27.
49. 4x*^  2x 4 1. 60. x^  Sxy + 9y2. 1. x'^ + 2xy^ y^. 2. 4x2 _
4xy + y^. 8. f»^42m+l. 4. l2w + m2. 6. 412 a HQa*.
6. 4x212xj/ + 9y«. 7. 9x*12x2y2 + 42^. 8. a* + 4a26+4 63.
9. 25  10 he + 6V. 10. 4 rt2x2 + 4 axy2 + j^*. n. a%'^  2 a6c + c^.
12. 49  28ax + 4a2x2. 13. c^ + (P + c^ + 2cd + 2c« + 2dc. 14. a^ + 6'
+ c2a6+2ac26c. 15. x^^y^^ z^ +2 xy 2 xz 2 yz. 16. l+c2+(P
+ 2c2rf2cd. 17. 9a2+ fc* + 1 6a6 + 6a26. 18. a2 + 46«
+ 1 4afe + 2a46. 19. x* + 4y^ + 9 4xy + 6x  12y. 20. a* +
4 52 _(_ 9cs H 4a6 + 6 ac + 12 6c. 21. 26 a^+^Hc^lO a610 ac+2 6c.
22. r7i'''+nHr^+s^+2mttf2»nr+2ms+2nr42ns+2r5. 28. m^+n^+r*
+ «2 _ 2 wn + 2 mr  2 w«  2 nr + 2 n«  2 rs. 24. a^ + b^ + c^ + I +
2 a6  2 ac  2 a  2 6c  2 6 + 2 c.
Page 94.26. x2  y^. 26. ??i*  n^, 27. 42/*  1. 28. 1  9aV.
29. 2630x + 9x2. 80. 49 a^ + 70 x + 25. 81. 16a* 462. 82. x*  y*.
88. x6y8. 84. aH3a_28. 86. x27ax+12a2. 86. x2+(6c)x6c.
87. x2 — (m — 71 )x — mn. 88. 9n2 + 7 w — 60. 89. wi2 f (r + »)w + rs.
40. x3+l. 41. x»l. 42. x*l. 48. a»+x». 44. 3fia\ 48. x^+B.
46. x827. 47. x*y*. 48. x»j/*. 49. x^xy\y^. 60. a2^.aa;rx*.
61. x8 + x2y + xy^ + y8. 62. «« _ ^2^. + aa;2 _ jgS. 88. a2 4. 4 „ 4. 16.
64. a83a* + 9a27. 66. x2  3. 86. 1  a2. 87. «« + a2 + a + 1.
68. x*+x8y+x22/2+ 3:^84.^4. 59. 4x22xy+j/2. go. 4a2+6a6+962.
61. 3 ax  4 62. 62. 2 mH  6. 68. a + 3. 64. x  4. 66. a2 _ 52.
66. x2  y*.
Page 96. —11. 5a8a;2^ 5a8x2. 12. 9x2y*, 9x2y*. 18. UmH*r*,
11 m2n<r8. 14. 7 wim2s6, 7 mn^s^. 18. 8 a68c2, 8 a68c2. 16. 12 xy2r2^ 12 xyz^.
Page 97. — 8. 5a6(2a+6). 4. 7 xyz(2x—9y), 8. ax(x2— axy+y2)^
6. a^h(a'^ab{b^). 7. o2(flfa;2 4. y _ i). g. 3a26(3a;  6y  1).
9. 6x8(3y2?/2 + 0). 10. 4x*y(l  3y  4y2 4. 2 y«). 11. 3x2y2
(X  2 x2 43 2/24 xy). 12. ay'^^l  3 a^ + 6 a^z^  aV^)
Page 98.— 2. (2x+y)(2x+2/) 8. (6 + x)(rt+y). 4. (ay)(a+x).
8. (2/6)(a + x). 6. (x2m)(yn). 7. (a + 6)(x2y2). g. (y'^+1)
(2/ + I). 9. (a6)(a24.3). 10. a (2/ 6) (x2 2/2). 11. (3a+2)(2a23).
12. (2a26)(3x22 2/2). 2. (2 x+2/)(2x+v). 8. (a+3 6)(a+3 6).
4. (a2_62)(a'i_52). 5. (3a;_2/)(3x2/). 6. (2x25 2/2)(2x25 j/2).
7. (xl)(xl).
Page99.— 8. (2?/l)(2?/l). 9. (3x4 v)(3x4 2/). 10. (5 x 2/2)
(5xv2). 11. (l2x26 2/)(12x25?/). 12. (1  5 a22/2)(l  6 a22/2).
13. (5 + 3a6)(6 + 3a6). 14. (210 rt6)(210 a6). 18. (xy^S)(xy^S).
16. (4 + 5a6^c3)(4 + 5a62c8). 17. (11 a + 100 6*)(ll a + 1006*).
18. (m2n20«2)(wi2w20s2). 19. (rt462)(a + 62). 20. (xyS)
(X2/3). 21. (m26n)(m26?0 28. (a + 6)(a  6)(x 4 2^)
(x\y). 24. (x + y{z)(x^y^ z). 26. {x  y \ z)(^x  y ^ z).
26. (a2_26 + 2)(a2_26 + 2).
" Page 100.— 4. (30254.4 c) (8 a26 4 c). 6. (x2/+2 2/«)(x2/2 2/?).
6. (x242/2)(x + 2/)(xv). 7. (3 v+l)(3 2/ 1). 8. (2ax+6a)(2(ix()a\
9. (4rt2x247 2)(4a2.y2_7;j). 10. (1 4l)02)(l 43^)(l 3;?). 11. (y'^^iz'^)
(y\2z){y2z). 12. {U^a'x*)(\\a^x'). 13. (x2+7 2/22f)(x27 ^2^).
14. (12 + 5rt2/)(125rt?/). 18. (x^^a~h)(x'^ a\b). 16. (a+Xy)
(ax + 2/) 17. (2x2 + a + 6)(2x2a6).
ANSWERS. 381
Page 101. — 18. 4xy. 19. (a + 6+c+(?)(a+6cd). 20. — 4mn.
21. (a6+c(?)(a6cfrf) 22. Sab{a^^b^). 23. (5al)(7a).
24. (5a6)(a + 56). 26. (Ox  y)(4a; + 6 2/). 29. (2 a  b \ xy)
(2abxy), 30. (2 x^ + Ssc  1)(2 a;'^  3x+ 1). 81. {Qabc)
l4abc). 32. (a  6 + 2 a:y)(a  6  2 xy). 88. (1  x + 4 aft^)
(lx4a62). 34. («  2 6 4 2x + 3y)(a  2 6  2x  3y). 36.
(a+l46c)(a+l6 + c). 36. (xy4w4»)(icy— w— 7i). 37.
{x \ y \ s — z){x\ y — 8 •}■ z), 38. (a — c + 6 + d)(a — c — 6 — d).
Pagel02. — 41. ix'^\y^^xy)(x^+y^.xy). 42. (x243?/2+2xy)
(x2 + 3 y2 _ 2 xy). 43. (m^ 2n^ + 2 mn) (m^2n^2 mil). 44.
(a* + ft2 4. a'^b){a* 4 6^  a^6). 45. (x*+3 y* + 2 x2y2)(a:4_f.3 y4_2 a;22/2^.
46. (2x23 2/2 4.2xy)(2x23y2_2xy).
Page 103. —48. (x2 + 2 + 2 x) (x2 + 2  2 x). 49. (8 + y2 + 4 y)
(8 + 2/24y). 60. (x + 4)x. 61. (x2 + 2 y2)x2. 62. (a2 + 2 62 + aft)
(a^^2b^ab). 63. {a^  b^ + 2ab)ia'^  b'^ 2ab).
Page 104.3. (x+5)(x42). 4. (xll)(x2). 5. (x9)(x5).
6. (x + 4)(x + 6). 7. (x+7)(xH8). 8. (x7)(x4). 11. (a; + 8)(x3).
12. (x12)(x + 5). 13. (x9)(x + 5). 14. (x+9)Cx7). 16. (x + 14)
(x4). 16. (x10)(x+7). 17. (x9)(x+8). 18. (x+7)(x6).
Page 105. — 19. (x + 4) (x + 4). 20. (x + 10) (x + 4). 21. (x + 9)
(x+7). 22. (x+12)(xf6). 23. (y+12)(y+8). 24. (x9 a)(x6a).
26. (x5a)(x3a). 26. (x7)(x8). 27. (x  1.3)(x 4). 28.
(y20)(y7). 29. (x+7)(x4). 30. (x49)(x2). 31. (x+11)'
(x6). 32. (x412)(x 11). 33. (x + 14)(x  13). 34. (xy  11)
(xy+2). 36. (xy8)(xy13). 36. (x12 a)(x+5 a). 37. (x11)
(x+9). 38. (217)(^+16). 39. (x425)(x+7). 40. (x + 16)(x6).
41. (x20)(x20). 42. (x + 13)(x  12). 43. (y4ll)(y + 6). 44.
(x19)(x4). 46. (x+3)(x18). 46. (y+14)(y3). 47. (y17)
(y+lO). 48. (y418)(y + 6). 49. (x2 + ll)(x2 f 6). 60. (x  2)
(x2 + 2x + 4)(x8'7). 61. (x4+ ll)(x2 + 3)(x23). 62. (y + 2)
(y22y+4)(yl)(y2+y4l). 63. (yS16)(y63). 64. (x+6)(x+c).
66. (x + a)(xc). 66. (y h a)(y  a){y + b)(y  b).
Page 106. — 68. (x16)(x+16). 69. (x416)(x+15). 60. (y+23)
(y13). 61. (y30)(y+ 16). 62. (x  27)(x  20). 63. (x  21)
(x+20). 64. (x+36)(x35). 66. (x+30)(x+8). 66. (x+40)(x33).
67. (xf24)(x23).
Page 107.4. (3x7)(x6). 6. (3x+6)(2x7). 6. (4m5)
(3m4). 7. (3xl)(x3). 8. (3x+l)(x2). 9. (5x12)(x7).
10. (3x+6)(x2 6). 11. (3x2y)(2x4y). 12. (6x4y)(x + 2 y).
13. (2r7i + 3mw)(mwn). 14. (3 a2&)(a+6). 16. 2(yo)(6ya).
16. (2xy)(2x43y). 17. (5x + 7 y)(3x  5y).
Page 108.20. (x+3)(3x+5). 21. (x42)(7 x+6). 22. (5x12)
Cx1). 23. (2x+3)(3x2). 24. (x2)(3x+6). 26. (3x8)(2x+6).
26. (2x3)(3x2). 27. (x f 3)(3x  4). 28. (x4)(3x6).
29. (x + 3)(10x7).
Page 109.2. (x + 5) (x + 4).
Page 110.— 6. (x+3)(3x4). 6. (x+3)(x+4). 7. (2x7)(x+4).
8. (3x7)Cx2). 9. (2x + TO)(x3»i). 10. (3x + n)(x 2n).
382 ANSWERS.
112.— 6. (a+2)(a;2_2a; + 4). 6. (xl)(x^{x{l). 7. (!+«)
(1  X + ar'*). 8. (X  '6){x^ 4 ^x f 9). 9. (5 + x)(25  5x2 ^ x*).
10. (4 + x^)(2 + x)(2  X). 11. (x2 + 16) (X + 4)(x  4). 12. (a^  3)
(«*+ 3aH 9). 13. (x2_ y)(x4+ x=^y + y'^) 14. (m 4 n)(w2 mn + n*)
(?/»  7i)(m2 f m/i 4 n^). 16. {m^ + »'^)(w<  mH^ 4 »*♦). 16. (x  1)
(x*4x8 4x2 + x+ 1). 17. (1 4a46)(l a6 + a24.2a6462).
18. ra  6 4 l)(a2 _ 2a6 + 6'^  a 4 6 4 1). 19. Smn(m^ + n^).
20. 2 n(3 m2 + n^. 21. 3(a 4 x)(a  x)(5 «« + 8 ax 4 5x^).
22. (a2x)(7aM8ax44x2). 23. (x + y + v^xy)(x + y V2xy) .
24. (a 4 b'^ + V2 «6^) (a 4 ft^ _ V2a62). 25. (?»2 + n^ + V 2 m^n* )
(m2 + n*  V2 wj2,i4)^ 26. (9 4 x2 + Vl8 x2) (9 4 x*  VlSx^) .
27. (a*46*4V2a*/>4)(a»46*\/2tf464). 28. (3 x2 + 2 y + Vl2x^)
(3x2 4 2yVl2x22/). 29. (2 Xj4 y^ 4 >/4x2/2)(2 x 4 2^^ _ V4xy2).
30. (x2 4 1« + V32X2) (x2 + 16  V32'x2) .
Page 115. — 7. (xl)(x4 2)(x43). 8. (x 4 !)(« + 2)(x 3).
9. (x42)(x43)(x5). 10. (x  l)(x  6)(x 4 7). 11. (x  2)
(x3)(x 5). 12. (xl)(x43)(x + 7). 13. (x  2)(x  2)(x + 3).
14. (X4 l)(x3)(3x4). 16. (x  l)(x  2)(2x  3). 16. (x4 1)
(6x26x4 13). 17. (x43)(2x26x+ 13). 18. (xl)(x2)
(x43)(x4). 19. (x42)(x2)(x2 4x44). 20. (x + 2)(x2)
(x2 3x41).
Page 116. — 1. (3xl)(3xl). 2.xy{x\ y)(x + y). 8.2x(x2)
(x2). 4. 5a2(x3yO(«32/'0 6. x(x6)(x5). 6. 12 ox(ax 4 6)
(ax — 4&).
Page 117. —7. 3a(mw)(wn). 8. 5a(a— 6)(a6). 9. (x'4l)
(X41). 10. 3a&(x22/2_8). 11. 7x2(x3v)(x32/). 12. a(a4l)
(a1). 13. {2xy{z)(2xyz). 14. 7x2(x3y+2xy^)(x3y2xy2).
16. bahy\x 4 ay)(x 4 ay). 16. 9(a2 4 2)(a2  2). 17. (x2 4 l)(x + 1)
(x1). 18. {a^^b'^){a^h){ah). 19. {x^+y^){x^\y^){x^y){xy).
20. (6+3x8)(53x8). 21. (l4w»2)(l4m)(lm). 22. {ah\cd)
{ahc\d). 23. 4m». 24. x(2x 4 3)(2x  3). 26. 5s(x2 + 3y2)
(x23y2). 26. 5(xv)(x+2/). 27. (,n^^h^\^a%^){n^^b^9a'^b'^).
28. (5x8)(x2). 29. 3x22/2(x+32/)(x3 2/). 80. 3ay(x3y)(x3y).
31. 5x(a 6 43x)(a63x). 32. (1 3x + 2xy2)(l  3x  2xi^2),
33. (x4d)(x4c). 34. (x+d)(xc). 35. (x— n)(x— wi). 36. {ad)
(6 4c). 37. (2cd)(a2 6). 38. (3x  y)(«  32:). 39. (x + 3)
Ixy). 40. (l4a6c2)(la4&c2). 41. (a2+ 024 ac) (024 c2ac).
42. (x2 + 1 + x) (x2 4 1  X). 43. (wi2 4. 7i2 4 mn) (m^ 4 n2  mn).
44. 4(x2 42 42x)(x2 422x). 45. 9(a2 + 2 4 2a)(a2 4 2  2a).
46. (w2 48n2 44win)(m2 4.8n24mn). 47. (x2 4 2 y2 + 2 xy)
(x242y2_2xy). 48. (2m2+n2+2m7i)(2m2+n22win). 49. (2a436)
(2a 36). 60. (x2 4 2^/2 + X2,)(x2 + 22/2 _ ^y). 61. (x2 4 V^ 4. xv)
(x2 4 2/2_x2/). 62. (a2 42 62)(a2_2 62). 53. (2 4 a2)(2  a^).
54. (lx)(l4a;4x2). 56. (22/)(442y4y2). 56. (xy)(x24xy4y2).
67. (a46)(a^a6 462). 58. (1 + a)(l  a 4 a^). 59. (x1)
(x2 + X 4 1). 60. (m 4 n)(w  n)(?7i2  mn 4 n^){m^ 4 wn + n*).
61. (2a2362)(4a*46a262 + 96*). 62. (3+x)(93x+x2). 63. (a242)
(a*  2 a2 4 4). 64. (a2 + 62) (^4 _ ^ifta + ^4). 65. (a 4 6)(a  6)
(a2 ah + 62) (aS + a6 + 62). 66. (a  x) (a* 4 a^x + a2x2 4 ax« 4 x*).
ANSWERS. 383
Page 118.— 67. {x\2)(x2)(ai^2x\4)(x^+2x^4). 68. (l + m)
(lm)(lm+w2)(l+w + 7»2). ^9. (;3_a:)(y+3x+a;i). 70. (l + x+y)
(1 xy^3(^^+2xy\y^). 71. (x  y + l)ix^ 2xy \ y^ x \ y ^ I),
72. (x««ll)(x+l)(xl). 78. (x'^7)(xfl)(xl). 74. (a+3)(a3)
(a4.1)(a_l). 76. (x+12)(x10). 76. (x+24)(x3). 77. (x+16)
(x15). 78. (x18)(x+8). 79. (x+17)(x+13). 80. (7x+l)(3xl).
81. (4x 7)(9x+8). 82. (x  2l)(x  20). 83. (x  15)(x  12).
84. x(x5)(x + 4). 85. (4a2)(a3). 86. (4a2 _ 2)(a2  2).
87. (3x+13)(x5). 88. r>(x^2xb). 89. (x7)(x+l). 90. (x5)
(x5). 91. (x7)(x + 2). 92. (x212)(x2+6). 93. (X'^lU)(X'^+4).
94. (3x+2y)(3x+2 2/)(3x22/)(3x22/). 96. (x+n)(xn)(x2+w2).
96. (x+a+l)(x+a+l) 97. {a{y){ay)(a^ay\y^). 98. (x23m)
(xH2). 99. (x + y2?)(x + 2/2). 100. (a 6 + cfd)(a 6 cd).
101. {mn^ab)(m—7ia^b). 102. (a2 + wn)(a— 2 — wi+w).
103. (3x2/+2s+22)(3xy2»2i?). 104. Cxy+l)(xy4l).
106. (ax  &)(x 4 y)(x + y). 106. (x2+ y^) (^^ ^. y) (a;  y) (a  6)(o 6).
107. (mn)(xy)(xy). 108. (xl)(x3)(x5). 109. (xl)(x+l)
(x + 3)(x5). 110. (x + 6)(x + 2)(x + l)(xl).
Page 121. — 6. 25. 7. 15 ab^c. 8. 8 xV »• ^ a*«V. 10. xy.
11. x1. 12. xHax2. 18. 15(xl). 14. 4(a262). 16. a^ab\b^,
16. x1. 17. x2 + x+l. 18. x24x + 4. 19. y + 2«. 20. x2 + y.
21. X + 4. 22. X  9. 28. X + 7. 24. x  12. 26. x + 1. 26. x2 + 12.
Page 122. — 27. x1. 28. x5. 29. 2 x + 3. 80. x  1.
Page 126.6. x5. 7. 2 x + 3. 8. x 4 5. 9. x  4. 10. x  3.
11. X 6. 12. X 4. 18. x29x + 21. 14. x2 + x+l. 16. x + 3.
16. X2X5. 17. x24x+3. 18. 2x5. 19. 2x2x43. 20. x^y^.
Page 128.— 4. Iba^y. 6. I2exh/^z^. 6. im a^b^<^c^x'^. 7. (x+y)
(xy)(x2fx2/+y2). 8. 35xy2(a+5)(a4 6)(a2a6+62). 9. a'^(ab)
(a4 6)(aHa6+62). 10. (x2)(x5)(x+5). 11. (x+10)(xll)
(x9)(2x5). 12. (x+5)(x+3)(xl). 13. (x12)(x+7)(x+5).
14. (7xl)(x5)(x+5). 16. a&(rt+6)(a6). 16. (x2f y2)(x+y)
(x+2/)(xy)(xy). 17. (xl)(x+l)(x2x+l)(x2+x+l).
Page 129.3. H.C.F., 2x3; L.C.M., (3x4)(12x24x21).
4. H.C.F., x5; L.C.M., x(x5)(x2+5x+2)(7 x2+x+l). 6. H.C.F.,
x+1; L.C.M., (a;+l)(3x2llxl)(6x217x7). 6. H.C.F.,xl;
L.C.M., (xl)(3x2 + 3x2)(21x2 + x10). 7. H.C.F., 4x3;
L.C.M., (4x3)(x2+4)(3x22x+4). 8. H.C.F., x+1; L.C.M.,
x2(x+l)(x2xl)(x2+x+l). 9. H.C.F., x26x+6; L.C.M., (x2)
(x3)(xl)(x4). 10. H.C.F., 2x3; L.C.M., (2x3)(3x2)
(x + 4)(3x + 4).
Page 132.1. ^. 2. ~ 8. ?^. 4. ^. 5. x + y.
36 a 3 m x + xy 3m2
a q8 4 a ^b + ab^ 4 6«
D. •
ab
Page 133.6. i^. 6. ■^. 7. l^m^. 8. 2i±2^
5 6 4 6x lip 3 6
3 4axy ^ jQ 2(a6), ^^ x+_2 ^^ _«±12_. ^3 a;49 ^
3x2 8y 9(a46) x9 x(x7) 4a(x4)
384 ANSWERS.
14. ^:^^^ 15, JLtl 16. gV12x+144 j^ x±a^
X  9 X + 20 x2 + xi/ + y2 x + 11 ' x + c
18. ?«. 19. ^ ^'tf . ' aO ^^ 21. ^. 22. ^'^+^^
xc x*+x*V+y* a^+x^ xa a(a+x)
28. ^^±1^ 24. ?^±^.
4(x + 6) 3x6
Page 134.26. ^]T^^^ , ae. 2£^. 27. _£JL5
88 «^ + 3a:2 gg x^43x+27 3^ x^6x+25^. 31 x2+3x7
x(4a+3x)" ' x2+3x+3' ' 2x23 x+ 15* * x2+3x+16*
82. ^'^^ . 88. ?!^l2x4:_5. 84. ^^.
x'^ 4 « + 1 x2 + X  2 X 4 y
Page 135. — 8. ax  b. 4. 6 + ?. 6. 6—. 6. x^y^.
a a
7. X + ^ — 8. x2 4 xy + 2^2. ^, a + b + =^. 10. x  2 
x{ y a — b X— 1
11. ax + ^. 12. x+2y^l^^. 18. x25ax+^. 14. 3+— ^•
a+x x\y 3x x2— 1
16. ab. 16. x2hx2/ + 2/^ ^.
xy
Page 136.6. ^^^M:^. 6. ^^^'^°. 7. ^i^. 8. 2o^Jzl«.
4/ t/ X "4" t/ X 1
a 2 ax ,rt 2a8aftfa2ft_262 a;3 x33xJ2^ + 3 0:22
S>.  • lU. ; • 11. • Is. •
a+x a\b x+y x — z
IS ^y 14 2^Llli!5?.
' x + y ' a4x
Page 138. 6. «i, A 4 6. A^, i, ^. 7. ^.
^ rtfec a6c a6c 12 ac 12 ac 12 ac 10x2
25 4x g 4q84a2& 3a2H3aft 2 62 ^ qxyf5xy ay\^y
10x2' 10x2* * 12 a8 ' 12a8 ' 12 a^" " x2y2 » 3.2^2 '
«^_+?.? 10 ^+ ^ 1 X 1 ji q2 _ 2 a& + 62 a + 6
x2t/2 * • X2  1' x2  1' X2  1* . ' d^  b^ ' CC^  b'^'
q8 + q62 4 q26 + 68 ^g ax 4 3a q2x42q2 2q
a262 ' ■x2 4"6x46' x2h6x + 6' x2 45x46*
,0 x8+3x23x9 3x + 3x2x39 x2  a2x24a2M2
id. . • — — • 14. : '^—l
x29 x29 x29 x*y*
62x2  62y2 ab
y\ j#4 /g4 yA.
Page 140.8. 1^. 4.^^^ + ^° 6.^=^ 6. «'2»y+y'
5x 2x2 x^ xy
 4a2x46q2_a;a ^ q6x 4 x2 ^ 2 x2y«  6 y* t 6 y^ I a^'
4q2x2 * ' a6y * * 10x2ya
10. y'^y^.
xy
ANSWERS. g85
Page 141.12. ^. 13. 2^,. 14. ^i^+i^. 15. 1«^.
^ x^y'^ a^b'^ x2_y2 cfib^
jg xy2y3x + 9 ^^ 5a;^13x + 8 ^g 8a ^^ 3(xl)
x2  6a; + 6 " ' x^ + a  12 * * 1  4 a^' ' aj2 _ 4 *
20. ^(7^+^). 21. t«^. 22. J. 23. «^(«±^. 24. ^.
jc2_2a;15 a^b'^oc^ ab a^  b^ xa
26. 4+^! 26. 0. 27. ^{^\^^^^') , 28. i^. 29. " ^
aa62 a2x2 a;^.y x^+ax+a^
30. A«i.. 81. y'^y . 82. ^'<^\ 83. «H«^6axHto^ ,
a^+fts x^+arV+y* ax{xy) ax(a^b'^)
84. a^ + «^>e5a + 606 3^ 2^y^. 3^ ^^^ 3^ 2(x + 1) ^
rt2  62 X*  y4 a
33 2a34.q62_^2q2ft_ft8 3Q Q ^ m8+mnHn8 ^^ 36
a2ft_ft8 • • • • (w+»)« ' * (x+3)2(x3)'
Page 142.42. ^^i±i^^. 43. ?+?«. 44. ^
a;2 + 2x15 a 3a lx
45 a^ + 9ax ^3 8 ^ x (x^Sx + S)
(x + a)2(ax)* (x216)(x3)' " (x2_ I6)(x6)(x 3)'
48. ^^^^^^ 49. '^
x310xM31x30 (a; + a)(x+ 6)(x + c)
Page 143.51. «^±^. 52, ^i^±J^^ 53. «^4a662
ax(a6) a262 a4_2a262 4.ft4
64. i 65. — 56. 1. 57. 0.
(a — c)(c — b) (x — a){a — b)
68 x2y + x y2 + 2xgg2y2g g^ 3 x^  a^  6^  c^
(x + 2/) (x  ^j) (y + 2) * (x — a) (x  6) (x  c)
Page 144.3. — 4. • 6. a  &. 6. — — 4. 5^^.
^ 6 x2 a 6 10^2
6. • 6. — • 7. a(x —
5 2 ^
Page 145.8. L. 9. — «^,. 10. ^ia±iL^.
a + x (a + 6)2 x2xy + y2
11. ?.
12. '^±y. 18. ^^ti.
X — 2/ X — y
Page 146.4. ^. 5. «^. 6.^^2^. 7. « • 8. «'
Say 2x2y2 2(xy) 3x a 6
9. x2  y2. 10. ? 11. xy. 12. x + 1. 13. £fizL^. 1. ^,
^ 3(6  c) a + 6 3
«>!».>. « 21x ^ 3rtv R 35ac2x2y2 ^ 662  ax ^ a
2. 4a6c. 3. — — 4. — •• 5. — mr^' 6. • 7. • 8. .
10a 2x ib^s ab a x
g 3Ca8 + a62 + a26 + &8) ^^ 2(x  y)2
6a ' 6y
whitb's alg. — 25
886 ANSWERS.
Page 147. U. ?i«^«^±^. 12. —T^ 7 18. 1* L
15. «  ^ 16. iBLlJ^. 17. _i 18. ^^^^^ 19. — !_
aO. (x+w)«. 21. ^ 22. ^±^. 28. ^^. 24. 1.
25. I ., ae. (^'  y'^)^ 27. — ^. 28. «^. 29. i^+5!.
80. x'^ + y^. 81. (a%2_i)2.
Page 149.4. H 6. ^. 6. .^ ^ 7. ^±^. 8. — 51±i— .
2/ x^+a;+l «— a ox+x— a
9. ''+"■ 10. a 1. 11. ''(«*). 12.x. 18. L. 14. "+6.
a a(J) — x) 1 + a b
18. «»«J'. ie.«i^!!zdi. „.k:i«^. 18. al. 19.?^±ii3?.
ax 6(<i'^l) l3o y"
" a«  6'^ o  1 a* + 1 a+b
«1+_L_. 8. "' + '''' . 7. 3a + 2 6. 8. ^+^. 9. i^.
a*Ha"^+l tt^ + ax + x" x + 3 2x + 3
10, ^?L. 11. ^J_. 12.1. 18. «. 14.^. 16. 2
a6 ax X a + 1 x'^— a^ a(a2— 1)
16. ^>«2 . 17. ini_, 18. i_. 19. ^y+^'y' + ^^y*.
ai/ x2 ^ 4 j/a x2  1 x*  y*
20 ^^>a^'^  f 30ax24aa g^ 8 q^x  4 ax' + 4 gS  a^ H x^ gg x^^
20 a(a + x) ' * ^a^x^ ' ' y^'
28. ^^^^^^ 24. 5f^. 26. «. 26. a(a2 + ft^). 27. ^^.
28. «'(^ + V).
a — h
Page 151.29. ^i^^t^. 80. <^fta;xy + x^ gj xy« + a;^^ + y^z^
10 a6y xy^
88. _?y_. 88. li(^±ii. 84. ,^.'» 85. . „^"' „ .
x'* — 2/*'* X wr — m^ — m m* 4 3 win + 2 n'
3g mHn?, 87.^^. 88.^^^:1^. 39. 2^(«zilVi«(«±l). 40. ^^^±^
2mn aH6^ ay+6a; 2/(a+l)x(al) 2w»a
41. 1. 42. 2^. 48. LzJMjx?. 44 2^^
a* f ic* 1 + a y^ + 2 xj/ — x*
Page 153. — 6. x=60. 6. x=3yV 7. x=6. 8. x=36. 9. x=12.
10. X = 0. 11. X = 7. 12. X = 9. 18. X = 12. 14. x = 9. 16. x = 5.
16. X = 10. 17. X = 4. 18. X = 6. 19. x = 8. 20. X = 1. 21. x = 1.
22. X = 1. 28. X = XO.
ANSWERS. 387
Page 154. — 84. x = 2. 25. x = 2f. 26. x = 3. 27. x = IJ.
. x = 4i. 29. x = 2. 30. x = 0. 31. x = 2. 32. x = 6.
33. X = IJ. 34. 2/ = 55. 35. x = 2. 36. x = 4. 37. x = 8. 38. x = 7.
39. x = l. 40. x = 2. 41. x = 4. 42. x = li. 43. x = 4. 44. x = 0.
45. X = f . 46. X = 2.
Page 155.  3. x = ^. 4. x = t^^— 5. Hrn + d ^ c)
a — b 3af26 a
e. x=«iftz:^. 7. x=2!=i*!. 8. :,= «»'(»+") . 9. :c ^°''''^''+''> 
a— c 26 w— a 4a6
10. x =  ^^/>'' 11. x = a6.
w + n 4 i> + »•
Page 156. — 12. x = a + b. 13. x = 14. x = ^ ~ ^^ ~ ^' .
* 6a + 26 a6
15. x = • 16. x= — : — 17. x= • 18. x =
6d b 26 — a a + 62c
19.x = «!^t^^±^. 20.x = ^?LziJ?. 21.x = i?L. 22.x=i>gr.
a+6 w+n 2n
1. 72 and 60. 2. ^^^ and J^^^^ 3. 24, 36, and 48.
n 4 w « 4 wj
Page 157. — 5. , ^, and • 6. First day, 27 mUes ; second, 36
6 3 2
miles; third, 24 miles. 7. Eldest, $3000; second, $ 2400 ; youngest,
$ 1800. 8. First, ^ 280 ; second, 1 2 10 ; third, $ 246. 9. Carriage, $ 88 ;
horse, $187. 10. Horse, $120; saddle, $20; bridle, $10. 11. B's age,
50 yrs. ; A's, 40 yrs. 12. Husband, 30 yrs. ; wife, 18 yrs. 13. 3y^j days.
Pagel58. — 15. 4 J days. 16. 1? days. 17. 6Jf days. (2) A, U^^V
days ; B, 18 A days ; C, 34^ days. 18. 24 days. 19. 24 days. 20. $56.
21. First, $3258; second, $2896.
Page 159. —22. Smaller, $3240 ; larger, $4860. 23. First, 60 yards ;
second,' 46 yards; third, 40 yards. 24. 11 ^ hours. 26. 20 days.
26. 60 sheep. 27. 36 persons, 60^ ; 18 persons, 76^ ; 46 persons, $ 1.60.
28. $ 1000. 29. Longest side, 36 rods ; second, 27 rods ; third, 18 rods.
30. 40 eggs.
Page 160. — 31. 10 and 11. 32. Men, $12; women, $9; children,
$4.50. 33. B, $1860; A, $2160; C$2000. 34. C, ^^^; A, ^; B,
^ ~ ^ ^ 35. 640 yards. 37. $ 46 and $ 76. 38. 76 and 60. 39. ^^
and
3 ., w . ^_^
bm
a— b
Page 161. — 40. 60 apples. 41. Pears, 40; lemons, 28; oranges, 8.
42. 8 children. 43. Barley, 18 bushels ; corn, 24 bushels ; oats, 30
bushels. 44. 1059 men. 45. Fivecent stamps, 9 ; twocent, 16 ; one
cent, 26. 46. 30 acres, 47. 21 days.
388
ANSWERS.
Page 162.— 48. ?L±_5?! days. 49. 4§} hours. 60. C, 40 days; A,
a + 6
120 days ; B, 60 days. 61. 50 yrs. 68. (1) lO^f minutes past 2 ; (2) 21^^
minutes past 4. 68. (1) 49^j minutes past 3 ; (2) 10}^ minutes past 8.
64. 1^1000. 66. 36 miles. 66. 6^ ^ouis, 67. 14^ miles.
Page 163. — 8. 4 p.m.
Page 164.— 4. 7.30a.m.
Page 165. — 8. First, 360 miles; second, 190 miles. 10. 12, 18, 5,
and 45. «
Page 166.— 11. 12, 28, and 60.
mn
18.
m
2 + n
fi,
m
2 + n
+ n, and
2f n
Page 167. — 16. 187 pounds. 17. 42%. 18. 4000 pounds.
Page 168.— 19. $28.94. 80. 8%. 81. Syr. 11 + mo. 88. $620.
83. 3yr. 9 mo. 84. $280.74+.
Page 172. — 6. x = 13 ; y = 5. 6. a = 8 ; y = 6. 7. x = 7 ; y = 5
8. X = 10 ; y = 6. 9. x = 24 ; y = 4. 10. x = 11 ; y = 5. 11. x = 8
y = 5. 12. X = 8 ; y = 6. 13. x = 12 ; y = 18. 14. x = 50 ; y = 20
16. X = 16 ; y = 12. 16. x = 3 ; y = 2.
Page 174. — 19. x = 5 ; y = 4. 80. x = 3 ; y = 7. 81. x = 7 ; y = 1
88. X = 6 ; y = 2. 88. y = 14 ; « = 10. 84. x = 11 ; y = 7. 86. x = 4
y = 2. 86. X = 6 ; 2? = 12. 87. x = 10 ; y = 9. 88. x = 6 ; y = 4.
Page 175.81. x = 12; y = 8. 38. x = 2 ; a? = 3. 88. y = 4 ; « = 5
84. X = 3 ; « = 2. 86. y = 5 ; 2: = 1. 86. x = J ; y = i^. 87. y = 14 ;
= 10. 88. X = 11 ; 2? = 13. 89. X = 16 ; y = 12. 40. x = 7 ; « = 3.
41. X = 12 ; y = 6. 48. x = — ; y = ?• 48. x = 16 ; y = 12.
a b
Page 176.1. x = 13; y = Q. 8. x = 2; y=3. 8. x = 11 ; y = 2.
4. X = 8 ; y = 4. 6. x = 4 ; y = 3. 6. x = 10 ; y = 5.
Page 177. — 7. x = 5 ; y = 6. 8. x = 8 ; y = 10. 9. x = 12 ; y = 9
10. x = l;y=2. 11. x = 6;y=12. 18. x=f;y = V. 18. x = 3
y=2. 14. x=J; y=l 16. x=7; y=3. 16. x=10; y=2. 17. x=38}
y = 70. 18. x = 6;y=12. 19. x = 11^ ; y = 4J. 80. x = 3;y = 5
81. x=9; y=123}. 88. x=d\', y=2j\, 88. x=5; y=8. 84. x=6
y = 4. 86. X = 13.1 ; y = 5.6. 86. x = 5 ; y = 3. 87. x = 4; y = 5
88. x=8; y=2. 89. x=^^i^^^; y^^kiiln^, 80. x=tt^ ;
^ 2
Page 178.81. x=^^±^ y=«^^. 88. ^(f"^^') ; ^("F^
* a+6 * a+6 m^  n^ '* ^
w
n2
33 ^_ mn(an\bm) , mn(ambn) ^ ^_ ll(a+6) . 9(a6)
m^hn^ to2 + n2 * ' 2 ' ^ 2 '
86. x =
86. x =
ab^
1 ' ' wj2»2_i a2+6^'
J^^^. 87. X = «i^^l26) K2a^L^. jg. ^^ = iiL» • « = !!.
ANSWERS. 389
Page 179. —40. x = 4; y = 6. 41. a = 4 ; y = 6. 42. x = 3 ; y = 5.
43. X = f ^ ~ ^% y=J*!.Zl^. 44. a;==3jy = 2. 45. « = 7 ; y = 6.
o» — am an — bm
46. as = ^^ ^^ ; y = ^^ — .^^« 47. « = ; y =
nq — mp * np — mq ' a + b* • a — b
Page 181. —6. x=S] y=2j e=l. 6. a;=7; y=5; 2j=4. 7. a;=3;
y = 4; z = 6, 8. x = 2; y = 3; 2r = 4. 9. x = 2; 2^ = 3; 2: = 7.
10. X = 2 ; y = 3 ; 2r = 6.
Page 182. —11. x = 5; y = 4; 2: = 3. 12. a; = 8 ; y = 4; z = 2,
13. x=ll; ?/ = 9; z = 4. 14. a; = 6; y = 12 ; 2? = 20. 15. x = 24 ;
y = 60; 2 = 120. 16. a; = 3 ; 2/ = 4 ; 2? = 6. 17. 3C=7J; y=7 ; ;?=lJ.
18. a;=y=g= ^ ^^^ » 19. a;=4— ; y= ^ ; «=t^
aft + ac+oc a+o— c a— 6+c b—a\c
20. sc = 4 ; 2/ = 9 ; 2; = 16 ; tJ = 25. 1. 147 and 196. 2. 65 and 24.
8. Boy's age, 18 yrs. ; girl's, 7 yrs.
Page 183.4. A's age, "^ + ^^^ ; B'sage, 9±^. 5. First, $2500;
m—n m—n
second, $7500. 6. Silver, 562 ounces ; copper, 60 ounces. 7. Nickel,
20 ounces; copper, 28 ounces; silver, 12 ounces. 8. 24. 9. A, $300;
B, $ 600. 10. Tea, 86 cents ; sugar, 8 cents. 11. Tea, ^^* ~ ^^ ; sugar,
9IL^^£m, 12. f.
ad — be
Page 184. — 18. if 14. y. 15. 1 and 6. 16. Upper, 45 inches ;
lower, 63 inches. 17. Finer, $1.20; coarser, $.80. 18. 10 and 2.
19. A's age, 49 yrs. ; B's age, 21 yrs. 20. Persons, 13 ; sum, $ 3.
Page 185. — 21. Larger, 10^ hours; smaller, 14 hours. 22. 9, 15,
and 24. 23. 276. 24. A, $3^ ; B, $3 ; C, $2^. 25. First, $8 ; second,
$ 18 ; third, $ 16. 26. Wheat, $ 1.26 ; rye, $ .95 ; oats, $ .80. 27. 2, 5, 7,
and 10. 28. First horse, $ 56 ; second, $ 33.
Page 187. — 3. 2^12. 4. y4«. 5. a;"'. 3. 2iSa^^ofiy^^. 4.^. 5. — .
Page 188.— 6. 64a^b\ 7. ^^. 8. 27 a^b^ 9. 25a*xV.
27 a^
10.  64 o866a^. 11.  aio66c6a;io. 12. 729 aisfti^co. 13.  8 a^^.
14. 81 a*b^^y^. 15. 0656^4. le. a^b^c\ 17. 2«a«'»ar"V"% when n is
even ; — 2'»a'*"a;«V% when n is odd. 18. ^'^''a^b'^x^, 19. x^y^z^,
20.  a2»+i62«+ic2n+i. 21.  27 y^'if^. 22. ^^^. 23.  A^^£.
^ 16 2^2 125 «»&«
24. ??^^. 25. «:^^: 26. «?^^. 27. ^^. 28.  ^— •
Page 190.1. oi^bx^yJt 10 a^j^  10 x22/8 + 5 ^.^^ _ 2^.
Page 191. — 2. a«  6 a^ft + 15 a*62 _ 20 a%^ + 15 a%^  6 afe^ 4. 56.
8. TO*  4 TO»n + 6 TO2n2  4 wm^ + n*. 4. a^ + 5 rt*x + 10 aH^ + 10 a^x^
+6ax*+x5. 5. a55a*x+10a8x210a2x345ax*x6. 6. x"'+7x62^
+ 21x62^ + 35x*2^ + 35xV + 21x22/« + 7x2^ + 2/^ 7. c*  4 c^d f 6 c^d*
4c(P + d*. 8. a^ + 6a66f 16a*62 + 20a868+ 15a26* + 6a6fi+6«.
8d0 ANSWERS.
9. l3a; + 3a«x». 10. x*  4x» + 6a5« 4x+ 1. 11. 56ii«6».
12. 120 c^cP, 18. Mofiy*. 14. a« 8 0^6+28 a*^* . 15. c»+9c»<l
+ 3a c^rf^ + .... 16. 2^0  10 a»y + 45 zV . 17. •• + 16 m^ii*
Qmn^^n^. 18. .. f 28a%»+8aa;7^a*. 19. 21 a^&^+Tafc^ft'.
Page 192. — 21. 27 o»  64 a^ft^ + 36 aM  8 6». 22. (fi9a*bc
+27 a262c227 6»c«. 28. a*M2a«68c+§a262c2_ja&c»4— • 94. — ia^ftc
16 16
+ i a'^b^c^  2 ab^c^ + Mc«. 25. 32 m^  80 m*n* + 80 iii«fi«  40 iii*ji»
+ 10 mn^ n^. 26. ^ 3 cC^b^c + 24 abc^  64 c».
Page 193. — 8. 25a*«V. 4. 216 a^b^e'^x^, 5. 32aWxiS!f»
6. 81 mi%Vy*. 7. p^zV^. 8. d^xiV^^ 9. —  10. ^^
tns 81 nV
11. — ^^ ^ . 12. —  — ^ „ ' 18. ^=^ — , when nis even: — .
when n is odd. 14. ^'*,^'' . 15. m* + 4 w»«n + 6 w^n^ 4. 4 „i«8 ^ „4.
16. m*6TO*n+10m8n210w2n8+6TOn*n6. 17. a»3a*6+3a2626»,
18. z*4a;«46 3c24z+l. 19. lSaj+Sx^x*. 20. l4z^^6g*
\7fi\r^' 21. a:»9x6a2+27«3a427a«. 22. yi2_9y8;j2^.27y42:4272«.
28. 16a* + l6a8a; + 6a%2 + ax8 + i^. 34. «?^ _ «!^ + aft^i _ c».
16 27 3
25. a;*4x2a+6a2i^+^. 26. a«4^+^4 27. xiH16xi*y
x^ X* a^ a* a»
+ 105xi8y2.. ... +106x2yi8+16xyiHy^*. 28. x^o  10 x^a^ + 45 x^o*
+ 45 x2ai«  10 xa>^ 4 a'^. 29. x^^ + 12 xS^y + 66 x^V 4. ... ^. ee x2y«>
+ 12 xy" + yi2. 80.  792 aV. 81.  252 x^. 82.  126 x*.
Page 194. — 1. 5 ax^. 2. 2 x^y^r*. 8. 4 m^nx*.
Page 195.— 4. ±5a6. 5. ±9rt2&8c4. e. ±4x2y. 7. 4a62c».
8. 3 77in2. 9. ±2a62a;*. 10. 6mn2»*. 11. ±Zxhj^z, 12. 3a2c*.
13. 2o2a;. 14. ab, 15. a«x2. 16. ±x«y». 17. x*y8. 18. ±^^
6xy2
19. ««^. SN>. 2^*. 21. ±1«^. 82. «7?. 88. *3f.
6 62c* x2y 4w8n* y^z «•
24. 
X*"y2n
a26"»
Page 196. — l.x2y. 2. 2x2+y. 8. x23y2. 4. 3 a 2 2/2. 5. 2al.
6. l + 3y8. 7. 3x. 8. ^y2. 9. x + y2. 10. x2+ 2y2 ^. s^g 4. 4.
Page 197. — 12. x2 2 xy + y2. ig. 3 a;2 2 xy + 5 y*. 14. a2 5 a  6.
15. 1  6 X + x2. 16. 2x  xy + J y.
Page 198. — 17. 1 2x + 3y. 18. 9  a + 6. 19. ix2+ y + .
20. Jxli/. 21. xy + . 22.Hx4y + «. 28. 1 x + x2 x8+ x*.
24. a»3a + ?^.
ANSWERS. g9l
Page 202. — 14. 263. 15. 307. 16. 240. 17. 459. 18. 702. 19. 723.
SO. 626.. 21.13.3. 22.6.72. 23. .094. 24. .025. 26.8.68+. 26.24.22+.
27. .238. 28. . 29. I6h SO. 32}. 81. .591 + . 32. .73+. 83. .
84. j\. 86. 1 .414 + . 36. l". 732 + . 87. 2.236 + . 88. 1 .870 + . 89. 2.569 + .
40. .948+. 41. .774+. 42. .816+. 48. .935+. 44. .845+. 46. .516+.
46. .741 + .
Page 203.— 8. x2 + 2a. 4. 2«a. 6. 5a8362. e. a^ + a + l.
7. x2 4.23c 4.
Page 204. —8. 2x'^x6. 9. l+3«3a;2. 10. x 11. al+.
X CL
12. 3a2  4a6  262. 13. 2x2 _ ^xy + y\ 14. 3x2  2ax + a\
Page 206. — 8. 551368. 9. 314432. 10. 456533. 11. 175616.
12. 1124864. 18. 1953125.
Page 207. — 16. 35. 17. 57. 18. 145. 19. 364. 20. 325. 21. 301.
22. 4.2. 23. 4.6. 24. .46. 26. 12.5. 26. 4.07. 27. 3.04. 28. \,
29. \%. 80. If. 81. 3^. 82. .712 + . 88. 1.908 nearly. 84. 4.
86. 1.259+. 86. 2.08+. 87. 2.15+. 88. .873+. 89. .941+. 40. .793+.
41. .427 + . 42. .464 + . 48. .411 + .
Page 208. — 44. 2 a + 3 6. 46. x2 2x  2. 46. x + y. 47. 1 + 2 a.
Page 211.1. y/cm. 2. V(x  y)^. 8. yl^ 4. Vcfiofi,
6. y /{x + y)2. 6. \{ax)J. 7^ \a"6«. 8. Va62. 9. Vx^.
laJIH 11.^. 12.4/1^11. u,V^, 14. A^.
^x + y 'X* ^(x — y)8 ' ah
Page 212.16. \^I6626, v''256, \/iOOO. 16. v^, v^,_v^.
^ n* 'n'» ^9 ^a ^h ^c
20. v^, v^xiy2, .jgy. 21. v^ (xy)8, v ^(x+y)2. 22. V(a6)2, V^T^T,
Va=^. 28. v^oiofts, ^^, ^ ^^\\ 1. 6\/3. 2. 4v^. 8. 4\/5.
^a* ^(g + 6)^ 
4. 3\/3. 6. 4a62v^. 6. \baby/bb^K
Page 213. — 7. 5aV6 a~2 6. 8. 2v^2xgy 233^ . 9. (a+6)Va^.
10. (xy)Vx2 + xy + y2. n. (x + y) v^x  y. 12. (x2y2)\/S.
13. 26(a2_25)V3. 14. 5a(ly)v^. 16. (a:2+xy)v^^. 16. JV2T.
6 2
17. i\^. 18. J v^. 19. v^. 20. 2aVl5x. 21.2av'l5a6. 22. "Vox.
28. — ^ — V5(a262). 24. _2«_ V3(a2  62). 26. J_Vx2^.
a262 J q 6 xy
26. (x + y)V^r^. 27. v'x^'^H:.
Page 214.— 28. VIO. 29. Vbax. 80. \/a23p. 31. (a+6)Va262.
82. v'(a+6)(a6)='. 88. \^2(a3 6). 84. yVx(xy). 86. v'a'«6(a+6).
1. \/8ax. 2. VOOtf^. 8. ^^. 4. ^^^^ 6. v^2ax. 6. \/9x^.
392 ANSWERS.
11. va  6. If. VI  a«. IS. y/{ab)b.
Page 215. — S. 7V2. 4. 17V6. 6. ISVS. 6. 7*5 V30. 7. W^^.
S. 19v2. 9. 7\/6. 10. 2OV0. 11. Jv^. 12. }V2.
Page 216. — IS. iV3. 14. Jv^. 16. Vv^. 16.  3v^l0.
17. iv4. IS. iva. IS. fl + l:l^^/3. 90. fV2. 21. ay(3a;6y) VxT
a \a h xj I
99. 0. 9S. 3((i»  ^)vW. 94. 2qHh0ax^ ^ q
96. (6a 5ac 2 aa:)>/a^^ "^
Page 217.4. 168. 6. 50\ /6. 6. f v^. 7. ^v^. 8. iV6. 9. JVlS.
10. 120 v^4. 11. 12 vl8. 19. \/337600. 18.6^^64. 14. 24v^. 16. v^.
16. ^yj^' 17. 12 aft v^. IS. 6xV5ay. 19. ISoxWdx. 90. Ca^cy.
21. ?^^V30. 99. eal^y/^^. 9S. 12V^^5^ 24. ahcx^Vii
10 ^ ax
96. a62^'864a5S^. 96. 16 aft v^x^  y*. 97. 5v^.
Page 218.99. x  Vx  6. M. 6x  8 V3x  24. 81. 6 + \/lO.
89. 2 a  Vab  3 6. 88. 16 a + syaft + 6. 84. xVx  yVy. 85. a"^ V3
+ 3a2_al. 86. 2^+6\/6 + 2\/4. 87. V2 4  VO  j v/3  3.
88. 4a 6. 89. x4y. 40.33. 41. ix12. 49. Vx*  21.
48. —X. 44. 2x.
Page 219.— 8. 6V2. 4. ^^Ve. 6. jVS. 6. jV3. 7. f. S. f.
9.2. 10. J\/36. 11. i\/l8. 19. iv/4320. 18. v^40. 14. 2v^5. 15. 6aVS'.
16. ^</^. 17. ^V^x, 18. A^. 19. ?^Vi^. 90. V^cT^f.
6x2 2x 4x 2x^
Page 220.— 1. fV3. 9. jVl6. 8. 6\^. 4. iV7. 6. 2V3. 6. iV2'.
7. 3^. g. _A_V3xy. 9. — \/2^. 10. AV6. 11. ^Vy. 19. Va6.
3xy 2a ^ y b
Page 221.18. 2V7. 14. 3V2. 16. 2. 16. 2V3. 17. J VS.
18. 3V22V3 jg TVV30. 20. ^^t^. 21. VIO + 3. 22. 3 VlO
6 3
3V3. 28. VIOVe: 24. I±^. 26. 6+2V6. 26. q+2V^ + 6 ^
2 a6
2^ 30+13V6 gg 6V70~2V2T g^ 6  V6 3^ 2~ V2+V6
19_ ' * 238 * * 5 * * 4
81 2 f V2 f VO 32 12 + 9V3 + 3 V5  6 VT5
4 ' 22
ANSWERS. 393
Page 222.36. .707 + . 86. 5.7735+. 87. 1.4433+. 88. .9622+.
89. 6.121 + . 40. .464 + . 41. .169 + . 42. .127 + . 48. 1.5892 + .
44. 1.6.32 + . 45. 6.854 + . 46. 2. 8. 16. 4^25V3. 6. 2V3. 6. 2v^.
7. 4v^. 8. x^ V^, 9 . a^W. V^. \^ a^h^yJ ah\ 11. a*3c*. \%,\^aWy^,
18. 256 (a  6) \/a  h. 14. x»Vy.
Page 223. —16. 2. 17. V3. 18. 5V2. 19. 5v^6. 20. yfa.
21. Vx. 22. ^yJa^hx. 28. ai\/a^. 24. v^o^x.
Page 224.3. \/6 + l. 4. jV38 + iV2. 6. 2 + ^3. 6. iy42
jy2. 7. 2V5 + 1._8. V6 + V2. 9. 3V2._ 10. VJ + Vy^, or
JV2 + JV3. 11. iV35iV5. 12. iV3 + jV2. 18. V5a + Va.
14. — r^ , or 3( V3  v^). 16. ^ , or \\(cM. + \/2).
V3 + v^ V34^V'2
Page 225. — 17. VS + v^. 18. 1 + \/5. 19. V6 + V6. 20. 3 \/3.
21. V5+VTI. 22. V3\/2.
Page 226. — 4. x=32. 5. x=9. 6. x=5. 7. x=25. 8. x=2.
9. x=16J. 10. x=6. 11. x=12. 12. x=26. 18. x=4. 14. x=5.
16. x=4. 16. x=27.
Page 227. —17. x=6a. 18. x=^^^. 19. x=0. 20. x=2.
2 h
21. X = a2  2a6 + h\ 22. x = ?^. 28. x = «1±^. 34. x = ^.
16 26 4
25. X = 2. ' 26. X = a. 27. x = Va2~=n. 28. x = «i^^li}!.
46
29. X = ^* ~.^^^ 80. x=10. 81. x = 4. 1. (} xy  J ax*) Vxy.
2. ^Vx2  y2. 8. (x+2a)\/a. 4. J\/6j{/^. 6. v^l08.
x+ y _ 6
6 y/ a^h 7 x^{xyx\/yyy/y g (6xax) Va6 q6(x2 1)
x^2/ ' ' b^ x^ab
9.x^y/¥^l. 10. v^^^y^^ . 11 x+2110v^34 ^^^^^
y x29 _ ^
18. V3 + 2v^. 14^fV2 + i>/J4. 16. i>/6+iV30. 16. 2V51.
 2V3 + 3\/2\/30 0 v^ + V22 ,^. 2\/6 on V7  3
17. • 18. 19. — —• 20. .
4 2 ^ _?
Page 228. — 21. 12\/l6200. 22. 2xyy^. 28. o*a2\/a66.
24. _J_Va262. 26. x = ± i\/3. 26. x = /f^^ 27. x= ^ ^^
a6 ^ 4 62+C4 c2+l
28. X = a — 1. 29. x = a + b. 80. x = 9.
Page 229.1. 2V^n^. 2. 5v^^. 8. 16V^. 4. 4ay/^^.
5. x*V^n;. 6. ?Vn. 7. — V^. 8. ^\/^. 9. a2x>/^.
2 4 5
10. (a+ 6)V1.
394 ANSWERS.
Page 230.— 3. WV^, 4. 2>/3T. 6. t\V^. 6. lOv'^n^.
7. 7V1. 8. jVi. 9. 2xy/^n, 10. V^^.
a
Page 231.— 8. VlG. 4.  15\/6. 5. 432. 6.  512V3.
7. J. 8. WVab. 9.  6\/^^. 10.  6 VT. 11. V^^.
12. iVlU. 18. V'6. 14. f. 15. V6. 16.  1.
Page 233. — 17. 17. 18. 7. 19. 2a2+6+6>/^. 30. 6(c2a2).
21. V'G2 + 2V3f 2V2. 22. f^. 28. x + J. 24. fts _ ^8.
25. 6 + 2 V6. 26. L^ii^^^^^. 27. l^v^. gg. «!:^^.
89.6 + 2V6. 80. ^ + ^^^ + ^ 3^ 6 + 10V32 ^ 82. i«P^.
a — 6 o3 a + x^
88. . (Q a;'^  2) V^ . 34^ 2v^^. 85. 2 + 3V^^^. 86. 76^^=^.
x'^ + 1
87. _7 + 4V3. 88. 30+10\/5. 89. 4a\/^. 40. z^2zyVzyy^,
41. 1.
Page 236. — 1. a*6*c. 2. 3a% 8. oift'c. 4. x^f/^z\ 5. 8a*6.
1 ^ ? xv2
6. a^x ^2/2. 7. 2 a"6«c. 8. x»2/»2r3. 9. ^bz'^. 10. aftVy*. 11. ^•
12. ^. 13. ab^c^^y^. 14. 6*xi. 15. ^. 16. a563. 17. ahfyi.
18. (27) J. 19. leJaaa;!. 20. at X a?. 21. at ^ at. 22. aJft'ic"!.
28. asftl + a'^l 24. (a*62^a^62)K 25. y/a^, 26. TiTT* ^ 1"
28.  ,. 29. J' 80. — . 81. Va. 32. — . 38.4. 84.64. 85.9.
Va y/a " c^Vb
86. I. 87. j/^. 88. 3. 89. 4. 40. y^V^* ^1 1^ ^ i 48. 13.
44. 8.
Page 238.5. a^  6 J. 6. a J  6"J. 7. x2+ 2 + aJ"^. 8. 2 + x"2j/i
+ x^'2/'. 9. xy. 10. xi1. 11. x242x^ + x4. 12. l + x*+x*.
18. a^ftK 14. a^aibs\bh 15. 9x"J6x"5y"i+4jr''. 16. xx3j/i
+ x^yty. 17. a4&. 18. z^^x^y^+y^. 19. 2a*6i 20. x'^+y"*.
21. x^2x*x*. 22. x2ixiyi + y2. 23. a'^ or a^. 24. S* 1.
25. 2/^. 26. a26"'2%i2 or — !^ 27. a«x«+*. 28. — or 6*. 3*
a^b 2 ci2 ^
Page 241.— 4. x=± 9. 5.x=±2. 6. x=±3. 7.x=±4. 8.x=±3.
9. x=±6. 10. x = ±7. 11. x:^±l. 12. x=±V^^. 18. x = ±7.
14. X = ± 2 v/5. 15. X = ± v^. 16. z = ± 3^/3T. 17. x = ± 3.
18. x = ±vTr. 19. x = ±2>/:rT. 20. x = ±V2. 21. x = ±^jjvl6i.
ANSWERS. 395
Page 242.88. x = ± f . 88. x = ± Vm\n. 84. x = 0. 85. x =
^ ,Vn^. 86. x = ±YV7. 87. x = j: ^ >/6^^4cA 88. x =
la2 " 62c
±iV62c*2a6c. 89.x=±i. 80. x=±5. 81. 12 and 20. 88. 6 and 21.
c
83. Width, 26.7+ rods ; length, 356+ rods. 84. Width, 24 rods ; length,
60 rods. 85. 10 and 6. 86. Father^ s age, 40 yrs. ; son^s, 10 yrs.
87. Vab a,nd Vab. 88. Son's age, Vabm ; father's, Vabm,
ha h a
Page 243. — 8. 7 and 2. 3. — 1 and 5. 4. 6 and 4. 5. } and — 2.
6. — J and J. 7. h and — c. 8. a and Vft. 9. — Va and Va.
Page 244. — 18. x = 4 or — 3. 13. x = 9 or — 5. 14. x = 7 or 3.
15. X = 8 or 4. 16. X = — 7 or 4. 17. x = — 8 or 7. 18. x = 14 or — 4.
19. x =  13 or 2. 80. x= 22 or 7. 81. x = 2 or 6. 88. x=f
or 5. 83. X = I or  6.
Page 245. — 84. x = — J or . 85. x = } or 1. 86. x =  or 5.
87. X = — a or — &. 88. x = a or 2 a. 89. x = a or 30. x = — Va
2
or  Vb, 31.  3. 38. 2 and 3. 33.  5 and 2. 34. 12 and  6.
35. — 1 and  4. 36. — 1 and 5.
Page246.— 39. x2 + 4x5 = 0. 40. x2 + 7x + 12 = 0. 41. 2x»
+ 3x2 = 0. 48. 12x2 a; 1 =0. 43, 6xa + 7x + 2 = 0. 44. »»
+ 4x = 0. 45. x25 = 0. 46. x23 = 0. 47. x2  4x+ 1 =0.
48. a;2+(a]x^' = 0.
Page 248. — 4. x = 5 or 3. 5. x = — 2 or — 10. 6. x = 3 or — 7.
7. X = 15 or — 3. 8. X = 19 or 1. 9. x = 4 or — 22. 10. x = 4 or 3.
11. x = 5orJ. 12. x = 4±V^. 13. x = 10 or  8f . 14. x = 2
or — 3. 15. X = — 5 or — 42. 16. x = 4 or 3. 17. x = 7. 18. x = 7 or
 Vt^ 10. X = 15 ± 4\/l4. 80. X = x^f ± A>/67 21 a; = ± 6.
82. X = 5 or  2. 23. x = ± 5.
Page 249. — 8. x = 4 or — . 3. x = 2 or J. 4. x = — 1 or — J.
5. x = 3 or 1.
Page 250. — 6. x = — J or — f . 7. x = 2 or  . 8. x = 9 or — ^.
9. X = 7 or  5. 10. X = 4 or — y. 11. x = 3 or J. 12. x = y^^ or — \.
13. X = Jjf or  f .
Page 251. — 16. x =  J or f . 16. x = 6 or  JJ. 17. x = 4f or — 8.
18. X = I or J. 19. X =  I or  ^. 20. x = J or . 21. x = 4 or — jj.
88. X = 2 or  \}. 23. x = 4 or  4. 24. x = 2 or  J.
Page 252.3. x = 37 or  13. 4. x = 48.2173 or  7.2173. 5. x = 17
or 7. 6. X = 4 or — J.
Page 253. —7. x = 7 or  26. 8. x = 6 or  J. 9. x = 2 or
6i. 10. x= ^or 4 . 11. x = 5 or J. 18. x = 1 ± jV^.
13. x = ^iL±:^i^Sj6m. 14. ^ = zl1±^^.
2m 2m
396 ANSWERS.
Page 254. — 4. x = a or 1. 6. a: =  or — 6. x = a* or 6*. 7. x = 2
1 rt b a
or. 8. ;J(liV5). 9. x = iiorl. 10. x = aor ^ —
a 2 ^ afl
11. X = a or 6. 12. x = ^aft ± Va^ 6^ + 4 a^fe*. 18. x = or (a+2).
14 X = a ± — 16. X = 2 a or — 6. 16. x = a or  (6 + c). 17. x = c
a
\d±Viicd2c^. 18. x = ^or ^A_. 19. x = TOori.
a f a + 6 m
Page 255. — 1. x = 17 or 4. 2. x = 14 or — 13. 8. x =  or — 1.
4. x = 2or. 6. x=§. 6. x = 6or6f 7. x = j ± ^^^ Vl3.
8. X = a or 2 a. 9. x = i or — 2. 10. x = J or — }. 11. x = — J or
f. 12. x = 0or4. 18. x=iorf. 14. x= for 2. 18. x = a
or  . 16. X = ±  V^. 17. X =  or a. 18. x = a ± Vft. 19. x = a
+ ft or 9lJlJ^, 20. X = 6 or . 21. x = 6 or  a. 22. x = c or — .
a \ b b a
28..x = 2a6or 6. 24. x = ^ ± J\/4a2 ^_ I2a + 1. 25. x = 2±v^.
26. X = 1 ± a/^. 27. X = 2a or  12. 28. x = § i W9Sa^.
29. X = 3 or J. 80. X = 5 or  ^%, 81. x = 36 or 12. 32. x = '^4 or 2.
Page 256.83. x = 6 or 21. 84. x =  V ± V^. 8 5. x = 4
^^ 1 9ft ^ >. ^^ 7 •« ^ 22i:2V3l9 ,a ^ 2 n ± V4 n^  3
or — 1. 86. X = 4 or X. 87. x = — == • 88. x = = .
^6 3
89. X = a or — 1. 40. x = 13 a or — 6 o. 41. x = a or — 2 a.
42. x = ± iVn2T4. 48. x = — ± — Vc^ ~4a6. 44. x=??^t^or
2 2a 2a m — n
2^=^. 46. x = w±n. 46. X = ^ ^ ^ » 47. x = 9ori. 48. x=lor
n f m 2c
L 49. x = 5i:2v^. 60. x=^. 61. x =  or J. 62. x = 9 or 5.
^ a{b
68. x = 0. 64. x=#. 66. x = cor. 66. x =± \/4o2 ^_ 9.
^ c 2 2
67. X = 10. 68. X = 4. 69. X = 6 or  V^. 60. x = 4 or 0. 61. x = 2
or A. 62. X = i ± tVv^3689. 63. x = ^5±^ or ^5^. 64. x = 8 or
V 66. x = ±V3. 66. x = Kl±V^)•
Page 259. — 6. X = ± 3 or ± 1. 7. x = ± 5 or ± J\/3. 8. x_=
±2>/2 or ±\/^. 9. x = 2 or \/^. 10. x = 2 or v^ 4.
11. X = Vp ± Vg + p2. 12. a; = ± \/8 ; ± 2 or ± V^H". 18. x = i 1 or
± \/f. 14.^ = 18 or 3. 16. x = 25 or 3. 16. x = ± y/2.
17. X = ?^ly^. 18. x= if or 0. 19. x=4 or v^. 20. x = 4 or  1.
5
21. x=3or v^^. 22. x=9or2; i(7±vT73). 28. x=l±2Vl6orl.
ANSWERS. 897
84. x=:l(9±y/'^ ^) or j(3zfcV 7). 26. «= 13 or 78. 26. x=f orO.
27. x=V2a^±2aV€^fb^+l>^. 28. x=±8or V(Y)8. 29. a;=±l.
80. a;=rt or J a.
Page 260. — 81. a; =5 , 2, and 3. 32. x = 2, 2, and — 3. 88. x =  1,
I, 2, and — 2 ; or x = ± 1 and ±2, 34. x = 2, — 3, 1, and 4.
Page 262.6. 12 and 5. 7. 60 and 12. 8. 29 and 21. 9. 45 and
62, or 52 and 45. 10. ^±iVm^4:n] ^TiVm24n.
2 ^
II. ^_.iVw2n2_4w/t; T— V^n^zriw^. 12. 9 or 11. 18. 18,
2 2n 2 2n
9, and 6.
Page 263. — 14. Persons, 19; amount each paid, $10. 15. Breadth,
34 rods ; length, 35 rods. 16. First square, 5625 men ; second square,
7225 men. 17. 4 miles per hour. 18. 10 sheep. 19. Age, 36 yrs. 20. A,
10 miles per hour ; B, 9 miles per hour. 21. 25 robes. 22. 118 sq. rds.
23. 63 acres ; selling price, $ 50 per acre.
Page 264. — 24. 15 pieces; cost of each, $84. 25. 12 persons.
26. $80. 27. 12 horses. 28. One pipe, 15 hours; other pipe, 10 hours.
15 miles per hour. 80. 600 lbs. 81. 84 sheep. 82. 20 cts.
per dozen ; x ^ z:ac±V^l±^abc^
2a
Page 266. — 2. x = 8 or — 5 ; y = 5 or — 8. 8. x = 5 or 12 J ; y = S
or  llj. 4. X = 8 or J ; y = 3 or  ][. 5. x = 5 or 13f ; y = 4 or  13J.
6. X = 5 or — 4 J ; y = 4 or — 3J f . 7. x = 1 or J ; y = 2 or f . 8. x = 4
or If? ; y = 5 or 8^. 9. x = 2 or  i ; y = 4 or f . 10. x = 1 or 2^y ;
y = 3 or 1^7. 11. X = 3 or  I3S5 ; y = 2 or  9^. 12. x = 5 or  1^^ ;
y = 6orlJ.
Page 267. —14. x=±3 or ±^V7 ; y=±2or ±\V1, 16. x=±l or
ip V2 ; y = ± 2 or ± j V2. 16. x = ± 3 or T ? Vf ; y = ± 1 or ± f \/7.
17. X = ± 4 or T f v/7 ; y = ± 2 or ± f \/7. 18. x = ± 6 or tW2j
y = i 5 or ± Y>/^ 10 «= ±3 or =F V21; y = ±2 or ± ?V21.
20. x=±4 or ±2V3; y=±S or ± V3. 21. x=±3 or ±§v^; y=±2
or ±\V2 . 22 . x=±3 or TJV^; y=zfc2 or ±jV2. 28. x=±8\/^
or db AV387I; y = ±V^ or ± iV387T.
Page 269. — 27. x = 7 or — 5 ; y = 5 or — 7. 28. x = 5 or — 3 ;
y = — 3 or 5. 29. x = 5 or — 6 ; y = — 6 or 5. 80. x = 3 or 2 ; y = 2
or 3. 81. X = 4 or — 3 ; y = f or — 2. 82. x = 5 or — 2 ; y = 2 or —5.
88. X = 4 or 3 ; y = 3 or 4. 84. x = 8 ; y = 5. 86. x = 5 or — 2 ; y = 2
or —5. 86. X = 4 or —3 ; y = 3 or —4. 87. x = 6 or —5 ; y = — 5 or 6.
88. X = 10 or — 5 ; y = 5 or —10.
Page 270. — 4. x = 5 ; y = 3. 8. x = 9 ; y = 7. 6. x = 3 or — 2 ;
y=2 or —3. 7. x=7 or 5; y=5 or 7. 8. x=±5; y=±3. 9. x=±6;
y = ±2.
Page 271. — 10. x = 5 or 3 ; y = 3 or 5. 11. x = 5 or  2 ; y =  2
or 6. 12. i«J = 2 or 1 ; y = 1 or 2. X8. ap = 3 or 2 ; y = 2 or 3 ; x =
i
398
4; y
y = 
aud 9.
Pat
13. L
pipe,
Pa
P?
S.x
P
3. r
9:r
9.
I
6.
C
4.
6.
ANSWER& 399
Faiec 29a— 1. 1030. S. 336. S. 3G9. 4. 35. S. 237.5. 6. 5050.
^ ^^
Pa^^ 299. — 8. N'. 9. (in?; »; (m8)2nl. 10.8=240. 11. d = J.
12. « = 143. 1S./=18.1. 14. a = 45. U. it=ll. iaa=3;5 = 6.
17. S444>. $48(j, $520, and $560. 18. 12 yards. 19. 140 days.
90. 297110 yards.
Page 30r5. 128. a ^^I,^.
Page 302.— 7. tsI^t 8. a = 6. 9. a = 15. 10. r = 3. 11. r=f.
12 2, J, 5, ?.S, If. 18. \ ^, ^, ^^ 1, V^, 3, 3v^, 9. 14. \ ?^,
*"*'' ^^ 9933 6 35
28 8V7 m^ 32vj ^4^ ^ a ^z _
245' 245' 1715' 1715* 1715* ' x h •'' ^^<^'
Page 304.— 90. #=29524. tl.s=\^4':l. 28.» = lfi. 83. a =3.
94. a = 1. 95. o = 8. 96. a = 979.2. 97. .' 98. 2. 29. li. 80. 3i.
81.
n1
Page 305.88. If. ^ 84.65; 85. ^\\. 86. ilg^J. 87. 8H.
88. 54^;^.
Page 306.1 r =3 ; #=4372. 8. «=635 ; r=2. 8. 0=5^; » = 1365iJ.
4. a = 2l; » = 40957J. 5. i = 1594323. 6. r = 2; n = 22. 7 « = 3.
8. « =T^. « » = 221?. 10 5. 1. I. y. V 11. r = tV or 10%.
12. 2, 10, 50. 18  3,  18,  108,  648, or 21,  12S, 77i,  462f .
14. $ 10,000. 15. 15,625,000,000 grains.
Page 307. — 16 891 1 feet ; 90 feet. 17. 3, 6, 12, 24, or  24,  12,
6, 3.
Page 308.— 8. 3. 4. ^. 5. V and V 6. J, A» A» 1. A.
Page 316. — 5. 4^95424. 6. 3.92942. 7. 3.80821. 8. 4.50651.
Page 317.— IL 3.65437. 12. 0.93725. 18. 1.89709. 14. 4.53866.
Page 318.— 15. 2, 3, 4, 7, 9. 16. 30, 500, 6000, .008, .09. 17. 12,
23, 45, 66, 84. 18. 136.249 ; 326.496 ; 597.493 ; 73.173 ; .93119.
19. 4375.05 ; 59621.9 ; 673.22 ; 8.16056 ; .00053274.
Page 323.6. 182.97 + . 7. 4.42+. 8. 6.0115+. 9. 66.394.
10. .00037658. 11. 6.0i:i8+. 12. .19114. 18. .000050962. 14. 19770.
15. 1.7018. 16. .0036648. 17. 1.7489+. 18. 1.011.
Page 324.— 19. n=8. 20. n=Q. 21. «=5. 22. n=5. 28. x=3.
24. X = 3. 25. X = IJ, nearly. 26. x = 1.029. 27. x = 1.59+.
28. j.^ ^ogc^og<»
log 6
Page 325.— 2. 8%. 8. $450. 4. n= ^^^^ —
^ /o V log(l+r)
Page 328.8. $8456.52. 4. $4856. 5. $3322. 6. 23.98 yis.
V. 23.44+ yrs. 8. $145985.20. 9. $1554.04. 10. $106.63.
11. $125. 12. $27272/].
400 ANSWERS.
Page 332.1. .J^ + L^. 2. §1 ??_. 3. ?4^i_
X — 6 ac — 8 X — 8 x — 7 x— 5 x—1
4. ^—1. 6. ^ + ^. 6. L + ^l. 7. 2 2
x+5 x2 3x+l 2x+l a+2 x+3 x a:*!
+ ^_. 8.3 + i_ + _!— ?. 9. « + ? !«+ 2
X — 1 X — 1 x+1 X X— 1 x + 1 2x — 1 2x f l'
10.^+??. ll.i^J_. 12.1+ 2 4
x3 x + 3 X x + 4 x + 2 5x~5 lOx+16'
Page 334. —1. 1  6x+ 15x2  45x8 + 136x* . 2. l + 2x + 3x2
+4x8+5x*+.... 3. lfx+5x2+13x8441x*4121x5+.... 4. l+2xfx2
x82x*x6f .... 6. 1 + ^x4 ]iV«^+ yVa^H ifi3«*H •••• 6. 1 + 3x
+ 7x2+ 17x8+41x*+99xs4 —. 7, 1 _ a; _ a;2+ 5x8 7x* x^H .
8. a  (2 a  6)x + (a  2 6)x2 + (4 a + 6)x8 _ (H a  4 6)x* + .....
Page338.4. 11.^+^^^ Ll^J* _^_ 5^^ 1
4 16 4.8.162 4.8.12.168 3.9
3 . 6 . 92 3 . 6 . 9 . 98 ' * 4.5 4 . 8 . 52 4 . 8 . 12 . 58
8. 1+^
a a2
2 2.4
7. 2^(l+i L:J— + ''^'^ V
\ 3.2.4 3.6.22.42^3.6.9.28.48 ;
I 3a8 f. ^ 2a 12 « 1.2.4 « ^^ ^g 0^41.
2.4.6 3 3.6 3.6.9
+6a5&2_i0a«&8+.... 11. af.?a"36+^a"*&2+?.lll§oV7^8+....
3 3.6 3.6.9
12.1+1.1 LJ_+ 14.9
5 5 5 . 10 . 52 5 . 10 . 15 . 58
Page339. — 2. 3.036589+. 8. 2.024398+. 4. 2.0022248+.
6. 3.0024648 + .
Page 343. —2. x = 2; y = l. S. x = &; y = 2. 4. x = l;y=l.
6. x = ^^ ^f ; y ^ a/  cd ^ g 1012=2. 7. 76. 8. 16.
ac — bd ac — hd
Page 347.9. x = 3; y = 2;z = l. 10. x = 6;y = 8;a = 10.
11. x = 2; y = 2; z = 0. 12. x=l; y = 2; z = Z.
Page 366.4. 120. 6. 5040. 6. 325. 7. 120 (at round table 24).
8. 6720. 9. 1.3699. 10. 34650.
Page368. — 3. 126. 4. 84. 6. 16 (16 incl.l). 6. 495. 7. 67.
6. 1,392,300. 9. 36. 10. 63 in A ; 62 in B.
■1
•; .
To avoid fine, this book should be returned on
or before the date last stamped below
tOH ••40
632282